What are the solutions to the equation 296=4x^2+140x ?

The solutions to the equation 296=4x^2+140x are x=2,x=-37

Find the zeros of 296=4x^2+140x

The zeros of 296=4x^2+140x are x=2,x=-37

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296 = 4 x^{2} + 140 x

x = 2, x = - 37

Solution steps

296 = 4 x^{2} + 140 x

Switch sides

4 x^{2} + 140 x = 296

Move

296

to the left side

4 x^{2} + 140 x - 296 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 140 \pm 14 0 ^{2} - 4 \cdot 4 ( - 296 )}{2 \cdot 4}

14 0^{2} - 4 \cdot 4 (- 296) = 156

x_{1, 2} = \frac{- 140 \pm 156}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 140 + 156}{2 \cdot 4}, x_{2} = \frac{- 140 - 156}{2 \cdot 4}

x = \frac{- 140 + 156}{2 \cdot 4} : 2

x = \frac{- 140 - 156}{2 \cdot 4} : - 37

The solutions to the quadratic equation are:

x = 2, x = - 37

C = 200 - 7 x + 0.345 x^{2}

x^{2} = 23 x

s^{2} + 18 s + 100

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

f - 1 (- 2) = f (f - 1 (- 2))

What are the solutions to the equation 296=4x^2+140x ?
The solutions to the equation 296=4x^2+140x are x=2,x=-37
Find the zeros of 296=4x^2+140x
The zeros of 296=4x^2+140x are x=2,x=-37