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8x^2+40x-288=0
a = 8; b = 40; c = -288;
Δ = b2-4ac
Δ = 402-4·8·(-288)
Δ = 10816
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{10816}=104$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-104}{2*8}=\frac{-144}{16} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+104}{2*8}=\frac{64}{16} =4 $
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