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2v^2-9v+19=v^2-10v+25
We move all terms to the left:
2v^2-9v+19-(v^2-10v+25)=0
We get rid of parentheses
2v^2-v^2-9v+10v-25+19=0
We add all the numbers together, and all the variables
v^2+v-6=0
a = 1; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·1·(-6)
Δ = 25
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{25}=5$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*1}=\frac{-6}{2} =-3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*1}=\frac{4}{2} =2 $
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