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2x^2+10x-48=0
a = 2; b = 10; c = -48;
Δ = b2-4ac
Δ = 102-4·2·(-48)
Δ = 484
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{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-22}{2*2}=\frac{-32}{4} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+22}{2*2}=\frac{12}{4} =3 $
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