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5x^2=4x+20
We move all terms to the left:
5x^2-(4x+20)=0
We get rid of parentheses
5x^2-4x-20=0
a = 5; b = -4; c = -20;
Δ = b2-4ac
Δ = -42-4·5·(-20)
Δ = 416
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{26}}{2*5}=\frac{4-4\sqrt{26}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{26}}{2*5}=\frac{4+4\sqrt{26}}{10} $
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