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4y^2-13=y2+14
We move all terms to the left:
4y^2-13-(y2+14)=0
We add all the numbers together, and all the variables
4y^2-(+y^2+14)-13=0
We get rid of parentheses
4y^2-y^2-14-13=0
We add all the numbers together, and all the variables
3y^2-27=0
a = 3; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·3·(-27)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:y_{1}=\frac{-b-\sqrt{\Delta}}{2a}y_{2}=\frac{-b+\sqrt{\Delta}}{2a}\sqrt{\Delta}=\sqrt{324}=18y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18}{2*3}=\frac{-18}{6} =-3y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18}{2*3}=\frac{18}{6} =3
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