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2x^2=10x
We move all terms to the left:
2x^2-(10x)=0
a = 2; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·2·0
Δ = 100
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*2}=\frac{0}{4} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*2}=\frac{20}{4} =5 $
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