Y2-5y+8=y2+7y-6

Solution for Y2-5y+8=y2+7y-6 equation:

2-5Y+8=Y2+7Y-6

We move all terms to the left:

2-5Y+8-(Y2+7Y-6)=0

We add all the numbers together, and all the variables

-(+Y^2+7Y-6)-5Y+2+8=0

We add all the numbers together, and all the variables

-(+Y^2+7Y-6)-5Y+10=0

We get rid of parentheses

-Y^2-7Y-5Y+6+10=0

We add all the numbers together, and all the variables

-1Y^2-12Y+16=0

a = -1; b = -12; c = +16;
Δ = b2-4ac
Δ = -122-4·(-1)·16
Δ = 208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{13}}{2*-1}=\frac{12-4\sqrt{13}}{-2}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{13}}{2*-1}=\frac{12+4\sqrt{13}}{-2}
$