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10b^2+15b-45=0
a = 10; b = 15; c = -45;
Δ = b2-4ac
Δ = 152-4·10·(-45)
Δ = 2025
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{2025}=45$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-45}{2*10}=\frac{-60}{20} =-3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+45}{2*10}=\frac{30}{20} =1+1/2 $
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