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3a^2-12a-15=0
a = 3; b = -12; c = -15;
Δ = b2-4ac
Δ = -122-4·3·(-15)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{324}=18$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-18}{2*3}=\frac{-6}{6} =-1 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+18}{2*3}=\frac{30}{6} =5 $
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