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15x^2+47x+36=0
a = 15; b = 47; c = +36;
Δ = b2-4ac
Δ = 472-4·15·36
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-7}{2*15}=\frac{-54}{30} =-1+4/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+7}{2*15}=\frac{-40}{30} =-1+1/3 $
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