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2b^2+14b-120=0
a = 2; b = 14; c = -120;
Δ = b2-4ac
Δ = 142-4·2·(-120)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1156}=34$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-34}{2*2}=\frac{-48}{4} =-12 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+34}{2*2}=\frac{20}{4} =5 $
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