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5x=2x^2-4
We move all terms to the left:
5x-(2x^2-4)=0
We get rid of parentheses
-2x^2+5x+4=0
a = -2; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·(-2)·4
Δ = 57
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{57}}{2*-2}=\frac{-5-\sqrt{57}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{57}}{2*-2}=\frac{-5+\sqrt{57}}{-4} $
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