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27x^2-23x=15
We move all terms to the left:
27x^2-23x-(15)=0
a = 27; b = -23; c = -15;
Δ = b2-4ac
Δ = -232-4·27·(-15)
Δ = 2149
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{-(-23)-\sqrt{2149}}{2*27}=\frac{23-\sqrt{2149}}{54} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{2149}}{2*27}=\frac{23+\sqrt{2149}}{54} $
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