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5x(2x-4)=22
We move all terms to the left:
5x(2x-4)-(22)=0
We multiply parentheses
10x^2-20x-22=0
a = 10; b = -20; c = -22;
Δ = b2-4ac
Δ = -202-4·10·(-22)
Δ = 1280
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-16\sqrt{5}}{2*10}=\frac{20-16\sqrt{5}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+16\sqrt{5}}{2*10}=\frac{20+16\sqrt{5}}{20} $
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