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8x^2+80x+96=0
a = 8; b = 80; c = +96;
Δ = b2-4ac
Δ = 802-4·8·96
Δ = 3328
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3328}=\sqrt{256*13}=\sqrt{256}*\sqrt{13}=16\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-16\sqrt{13}}{2*8}=\frac{-80-16\sqrt{13}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+16\sqrt{13}}{2*8}=\frac{-80+16\sqrt{13}}{16} $
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