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5x^2-20x=12
We move all terms to the left:
5x^2-20x-(12)=0
a = 5; b = -20; c = -12;
Δ = b2-4ac
Δ = -202-4·5·(-12)
Δ = 640
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{10}}{2*5}=\frac{20-8\sqrt{10}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{10}}{2*5}=\frac{20+8\sqrt{10}}{10} $
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