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2x^2-20x=18
We move all terms to the left:
2x^2-20x-(18)=0
a = 2; b = -20; c = -18;
Δ = b2-4ac
Δ = -202-4·2·(-18)
Δ = 544
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{34}}{2*2}=\frac{20-4\sqrt{34}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{34}}{2*2}=\frac{20+4\sqrt{34}}{4} $
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