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5b^2+24b-36=0
a = 5; b = 24; c = -36;
Δ = b2-4ac
Δ = 242-4·5·(-36)
Δ = 1296
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{1296}=36$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-36}{2*5}=\frac{-60}{10} =-6 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+36}{2*5}=\frac{12}{10} =1+1/5 $
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