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50=2x^2+5x
We move all terms to the left:
50-(2x^2+5x)=0
We get rid of parentheses
-2x^2-5x+50=0
a = -2; b = -5; c = +50;
Δ = b2-4ac
Δ = -52-4·(-2)·50
Δ = 425
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{17}}{2*-2}=\frac{5-5\sqrt{17}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{17}}{2*-2}=\frac{5+5\sqrt{17}}{-4} $
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