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=12V^2-200V+600
We move all terms to the left:
-(12V^2-200V+600)=0
We get rid of parentheses
-12V^2+200V-600=0
a = -12; b = 200; c = -600;
Δ = b2-4ac
Δ = 2002-4·(-12)·(-600)
Δ = 11200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{11200}=\sqrt{1600*7}=\sqrt{1600}*\sqrt{7}=40\sqrt{7}$$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-40\sqrt{7}}{2*-12}=\frac{-200-40\sqrt{7}}{-24} $$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+40\sqrt{7}}{2*-12}=\frac{-200+40\sqrt{7}}{-24} $
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