What are the solutions to the equation 24=-16x^2+40x ?

The solutions to the equation 24=-16x^2+40x are x=1,x= 3/2

Find the zeros of 24=-16x^2+40x

The zeros of 24=-16x^2+40x are x=1,x= 3/2

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24 = - 16 x^{2} + 40 x

x = 1, x = \frac{3}{2}

Decimal Notation

x = 1, x = 1.5

Solution steps

24 = - 16 x^{2} + 40 x

Switch sides

- 16 x^{2} + 40 x = 24

Move

24

to the left side

- 16 x^{2} + 40 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 16 ) ( - 24 )}{2 ( - 16 )}

4 0^{2} - 4 (- 16) (- 24) = 8

x_{1, 2} = \frac{- 40 \pm 8}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- 40 + 8}{2 ( - 16 )}, x_{2} = \frac{- 40 - 8}{2 ( - 16 )}

x = \frac{- 40 + 8}{2 ( - 16 )} : 1

x = \frac{- 40 - 8}{2 ( - 16 )} : \frac{3}{2}

The solutions to the quadratic equation are:

x = 1, x = \frac{3}{2}

v^{2} + 4 = 85

p, [(p \land q) \lor \neg (p \land q)] = T

x^{2} + 4 = 8 x + 5

4 x^{2} = 12 x + 26

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

What are the solutions to the equation 24=-16x^2+40x ?
The solutions to the equation 24=-16x^2+40x are x=1,x= 3/2
Find the zeros of 24=-16x^2+40x
The zeros of 24=-16x^2+40x are x=1,x= 3/2