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2v^2+7v-23=7
We move all terms to the left:
2v^2+7v-23-(7)=0
We add all the numbers together, and all the variables
2v^2+7v-30=0
a = 2; b = 7; c = -30;
Δ = b2-4ac
Δ = 72-4·2·(-30)
Δ = 289
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{289}=17$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*2}=\frac{-24}{4} =-6 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*2}=\frac{10}{4} =2+1/2 $
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