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-8=4x^2+16x
We move all terms to the left:
-8-(4x^2+16x)=0
We get rid of parentheses
-4x^2-16x-8=0
a = -4; b = -16; c = -8;
Δ = b2-4ac
Δ = -162-4·(-4)·(-8)
Δ = 128
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{2}}{2*-4}=\frac{16-8\sqrt{2}}{-8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{2}}{2*-4}=\frac{16+8\sqrt{2}}{-8} $
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