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5b^2+25b-120=0
a = 5; b = 25; c = -120;
Δ = b2-4ac
Δ = 252-4·5·(-120)
Δ = 3025
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{3025}=55$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-55}{2*5}=\frac{-80}{10} =-8 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+55}{2*5}=\frac{30}{10} =3 $
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