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r^2-10r=50
We move all terms to the left:
r^2-10r-(50)=0
a = 1; b = -10; c = -50;
Δ = b2-4ac
Δ = -102-4·1·(-50)
Δ = 300
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{3}}{2*1}=\frac{10-10\sqrt{3}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{3}}{2*1}=\frac{10+10\sqrt{3}}{2} $
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