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15x^2+97x+56=0
a = 15; b = 97; c = +56;
Δ = b2-4ac
Δ = 972-4·15·56
Δ = 6049
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(97)-\sqrt{6049}}{2*15}=\frac{-97-\sqrt{6049}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(97)+\sqrt{6049}}{2*15}=\frac{-97+\sqrt{6049}}{30} $
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