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5a^2+20a+15=0
a = 5; b = 20; c = +15;
Δ = b2-4ac
Δ = 202-4·5·15
Δ = 100
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{100}=10$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10}{2*5}=\frac{-30}{10} =-3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10}{2*5}=\frac{-10}{10} =-1 $
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