What are the solutions to the equation 2x^2-3x-36=0 ?

The solutions to the equation 2x^2-3x-36=0 are x=(3+3sqrt(33))/4 ,x=(3-3sqrt(33))/4

Find the zeros of 2x^2-3x-36=0

The zeros of 2x^2-3x-36=0 are x=(3+3sqrt(33))/4 ,x=(3-3sqrt(33))/4

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2 x^{2} - 3 x - 36 = 0

x = \frac{3 + 3 33}{4}, x = \frac{3 - 3 33}{4}

Decimal Notation

x = 5.05842 \dots, x = - 3.55842 \dots

Solution steps

2 x^{2} - 3 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 36 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 36) = 333

x_{1, 2} = \frac{- ( - 3 ) \pm 3 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 3 33}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 3 33}{2 \cdot 2}

x = \frac{- ( - 3 ) + 3 33}{2 \cdot 2} : \frac{3 + 3 33}{4}

x = \frac{- ( - 3 ) - 3 33}{2 \cdot 2} : \frac{3 - 3 33}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 3 33}{4}, x = \frac{3 - 3 33}{4}

5 x^{2} = 256

25 = 15 x - 1.86 x^{2}

9 x^{2} - 11 x = 6 x

y, x^{2} + 2 x y + 4 y^{2} = 12

8 p^{2} + 14 p - 9 = 6

What are the solutions to the equation 2x^2-3x-36=0 ?
The solutions to the equation 2x^2-3x-36=0 are x=(3+3sqrt(33))/4 ,x=(3-3sqrt(33))/4
Find the zeros of 2x^2-3x-36=0
The zeros of 2x^2-3x-36=0 are x=(3+3sqrt(33))/4 ,x=(3-3sqrt(33))/4