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How do you solve 2b2 plus 5b-35? - Answers

Assuming the equation is 2b2+5b-35=0, the quadratic formula is one way to solve this equation. It can't be factored easily. Do not confuse the "b" variable in the formula given with the "b" in the related link. The values of a, b, & c are as follows to be used in the quadratic formula: a=2, b=5, c= -35. Placing these values in the quadratic formula gives: (-5+/- sqrt(52-4*2*(-35)))/(2*2). This yields variable b = (-5+/- sqrt(305))/4 or (-5+17.46425)/4 & (-5-17.46425)/4 = 3.11606 & -5.61606 which are the two roots of the given equation.



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How do you solve 2b2 plus 5b-35? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_solve_2b2_plus_5b-35

Assuming the equation is 2b2+5b-35=0, the quadratic formula is one way to solve this equation. It can't be factored easily. Do not confuse the "b" variable in the formula given with the "b" in the related link. The values of a, b, & c are as follows to be used in the quadratic formula: a=2, b=5, c= -35. Placing these values in the quadratic formula gives: (-5+/- sqrt(52-4*2*(-35)))/(2*2). This yields variable b = (-5+/- sqrt(305))/4 or (-5+17.46425)/4 & (-5-17.46425)/4 = 3.11606 & -5.61606 which are the two roots of the given equation.



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https://math.answers.com/math-and-arithmetic/How_do_you_solve_2b2_plus_5b-35

How do you solve 2b2 plus 5b-35? - Answers

Assuming the equation is 2b2+5b-35=0, the quadratic formula is one way to solve this equation. It can't be factored easily. Do not confuse the "b" variable in the formula given with the "b" in the related link. The values of a, b, & c are as follows to be used in the quadratic formula: a=2, b=5, c= -35. Placing these values in the quadratic formula gives: (-5+/- sqrt(52-4*2*(-35)))/(2*2). This yields variable b = (-5+/- sqrt(305))/4 or (-5+17.46425)/4 & (-5-17.46425)/4 = 3.11606 & -5.61606 which are the two roots of the given equation.

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      Assuming the equation is 2b2+5b-35=0, the quadratic formula is one way to solve this equation. It can't be factored easily. Do not confuse the "b" variable in the formula given with the "b" in the related link. The values of a, b, & c are as follows to be used in the quadratic formula: a=2, b=5, c= -35. Placing these values in the quadratic formula gives: (-5+/- sqrt(52-4*2*(-35)))/(2*2). This yields variable b = (-5+/- sqrt(305))/4 or (-5+17.46425)/4 & (-5-17.46425)/4 = 3.11606 & -5.61606 which are the two roots of the given equation.

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