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8x^2+16x-48=0
a = 8; b = 16; c = -48;
Δ = b2-4ac
Δ = 162-4·8·(-48)
Δ = 1792
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{7}}{2*8}=\frac{-16-16\sqrt{7}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{7}}{2*8}=\frac{-16+16\sqrt{7}}{16} $
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