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2=2x^2+5x
We move all terms to the left:
2-(2x^2+5x)=0
We get rid of parentheses
-2x^2-5x+2=0
a = -2; b = -5; c = +2;
Δ = b2-4ac
Δ = -52-4·(-2)·2
Δ = 41
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{41}}{2*-2}=\frac{5-\sqrt{41}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{41}}{2*-2}=\frac{5+\sqrt{41}}{-4} $
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