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X(2X-3)=15
We move all terms to the left:
X(2X-3)-(15)=0
We multiply parentheses
2X^2-3X-15=0
a = 2; b = -3; c = -15;
Δ = b2-4ac
Δ = -32-4·2·(-15)
Δ = 129
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{-(-3)-\sqrt{129}}{2*2}=\frac{3-\sqrt{129}}{4} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{129}}{2*2}=\frac{3+\sqrt{129}}{4} $
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