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125=10x^2-5x
We move all terms to the left:
125-(10x^2-5x)=0
We get rid of parentheses
-10x^2+5x+125=0
a = -10; b = 5; c = +125;
Δ = b2-4ac
Δ = 52-4·(-10)·125
Δ = 5025
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5025}=\sqrt{25*201}=\sqrt{25}*\sqrt{201}=5\sqrt{201}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{201}}{2*-10}=\frac{-5-5\sqrt{201}}{-20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{201}}{2*-10}=\frac{-5+5\sqrt{201}}{-20} $
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