24=(8x^2)-28x

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Solution for 24=(8x^2)-28x equation:



24=(8x^2)-28x
We move all terms to the left:
24-((8x^2)-28x)=0
We get rid of parentheses
-8x^2+28x+24=0
a = -8; b = 28; c = +24;
Δ = b2-4ac
Δ = 282-4·(-8)·24
Δ = 1552
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{97}}{2*-8}=\frac{-28-4\sqrt{97}}{-16} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{97}}{2*-8}=\frac{-28+4\sqrt{97}}{-16} $

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