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2x^2+20x-48=0
a = 2; b = 20; c = -48;
Δ = b2-4ac
Δ = 202-4·2·(-48)
Δ = 784
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{784}=28$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-28}{2*2}=\frac{-48}{4} =-12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+28}{2*2}=\frac{8}{4} =2 $
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