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2x^2-40x+192=0
a = 2; b = -40; c = +192;
Δ = b2-4ac
Δ = -402-4·2·192
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8}{2*2}=\frac{32}{4} =8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8}{2*2}=\frac{48}{4} =12 $
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