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x^2-10x=300
We move all terms to the left:
x^2-10x-(300)=0
a = 1; b = -10; c = -300;
Δ = b2-4ac
Δ = -102-4·1·(-300)
Δ = 1300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{13}}{2*1}=\frac{10-10\sqrt{13}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{13}}{2*1}=\frac{10+10\sqrt{13}}{2} $
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