What are the solutions to the equation -4x^2-7x+36=0 ?

The solutions to the equation -4x^2-7x+36=0 are x=-4,x= 9/4

Find the zeros of -4x^2-7x+36=0

The zeros of -4x^2-7x+36=0 are x=-4,x= 9/4

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- 4 x^{2} - 7 x + 36 = 0

x = - 4, x = \frac{9}{4}

Decimal Notation

x = - 4, x = 2.25

Solution steps

- 4 x^{2} - 7 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 4 ) \cdot 36}{2 ( - 4 )}

(- 7)^{2} - 4 (- 4) \cdot 36 = 25

x_{1, 2} = \frac{- ( - 7 ) \pm 25}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 25}{2 ( - 4 )}, x_{2} = \frac{- ( - 7 ) - 25}{2 ( - 4 )}

x = \frac{- ( - 7 ) + 25}{2 ( - 4 )} : - 4

x = \frac{- ( - 7 ) - 25}{2 ( - 4 )} : \frac{9}{4}

The solutions to the quadratic equation are:

x = - 4, x = \frac{9}{4}

4 x^{2} + 23 x = - 28

2 x^{2} + 3 x + 0.5 = 0

2 (x - 7)^{2} - 31 = 11

0 = 50 + 20 \cdot t - 4.9 \cdot t^{2}

(x + 1)^{2} = (2 x - 3)^{2}

What are the solutions to the equation -4x^2-7x+36=0 ?
The solutions to the equation -4x^2-7x+36=0 are x=-4,x= 9/4
Find the zeros of -4x^2-7x+36=0
The zeros of -4x^2-7x+36=0 are x=-4,x= 9/4