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8x^2+40x+48=0
a = 8; b = 40; c = +48;
Δ = b2-4ac
Δ = 402-4·8·48
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8}{2*8}=\frac{-48}{16} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8}{2*8}=\frac{-32}{16} =-2 $
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