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2x(5x-9)=27
We move all terms to the left:
2x(5x-9)-(27)=0
We multiply parentheses
10x^2-18x-27=0
a = 10; b = -18; c = -27;
Δ = b2-4ac
Δ = -182-4·10·(-27)
Δ = 1404
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1404}=\sqrt{36*39}=\sqrt{36}*\sqrt{39}=6\sqrt{39}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{39}}{2*10}=\frac{18-6\sqrt{39}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{39}}{2*10}=\frac{18+6\sqrt{39}}{20} $
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