7=(-16x^2)+32x

Solution for 7=(-16x^2)+32x equation:

7=(-16x^2)+32x

We move all terms to the left:

7-((-16x^2)+32x)=0


We calculate terms in parentheses: -((-16x^2)+32x), so:

(-16x^2)+32x

We get rid of parentheses

-16x^2+32x

Back to the equation:

-(-16x^2+32x)


We get rid of parentheses

16x^2-32x+7=0

a = 16; b = -32; c = +7;
Δ = b2-4ac
Δ = -322-4·16·7
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-24}{2*16}=\frac{8}{32}
=1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+24}{2*16}=\frac{56}{32}
=1+3/4
$