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4x^2+28x-40=8
We move all terms to the left:
4x^2+28x-40-(8)=0
We add all the numbers together, and all the variables
4x^2+28x-48=0
a = 4; b = 28; c = -48;
Δ = b2-4ac
Δ = 282-4·4·(-48)
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*4}=\frac{-28-4\sqrt{97}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{97}}{2*4}=\frac{-28+4\sqrt{97}}{8} $
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