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=(15-2V)(8-2V)
We move all terms to the left:
-((15-2V)(8-2V))=0
We add all the numbers together, and all the variables
-((-2V+15)(-2V+8))=0
We multiply parentheses ..
-((+4V^2-16V-30V+120))=0
We calculate terms in parentheses: -((+4V^2-16V-30V+120)), so:We get rid of parentheses
(+4V^2-16V-30V+120)
We get rid of parentheses
4V^2-16V-30V+120
We add all the numbers together, and all the variables
4V^2-46V+120
Back to the equation:
-(4V^2-46V+120)
-4V^2+46V-120=0
a = -4; b = 46; c = -120;
Δ = b2-4ac
Δ = 462-4·(-4)·(-120)
Δ = 196
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}$$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(46)-14}{2*-4}=\frac{-60}{-8} =7+1/2 $$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(46)+14}{2*-4}=\frac{-32}{-8} =+4 $
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