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x^2-4x=18=5x-1
We move all terms to the left:
x^2-4x-(18)=0
a = 1; b = -4; c = -18;
Δ = b2-4ac
Δ = -42-4·1·(-18)
Δ = 88
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{22}}{2*1}=\frac{4-2\sqrt{22}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{22}}{2*1}=\frac{4+2\sqrt{22}}{2} $
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