9+y^=(y-4)(y-1)

Solution for 9+y^=(y-4)(y-1) equation:

9+y^=(y-4)(y-1)

We move all terms to the left:

9+y^-((y-4)(y-1))=0

We add all the numbers together, and all the variables

y-((y-4)(y-1))+9=0

We multiply parentheses ..

-((+y^2-1y-4y+4))+y+9=0


We calculate terms in parentheses: -((+y^2-1y-4y+4)), so:

(+y^2-1y-4y+4)

We get rid of parentheses

y^2-1y-4y+4

We add all the numbers together, and all the variables

y^2-5y+4

Back to the equation:

-(y^2-5y+4)


We add all the numbers together, and all the variables

y-(y^2-5y+4)+9=0

We get rid of parentheses

-y^2+y+5y-4+9=0

We add all the numbers together, and all the variables

-1y^2+6y+5=0

a = -1; b = 6; c = +5;
Δ = b2-4ac
Δ = 62-4·(-1)·5
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{14}}{2*-1}=\frac{-6-2\sqrt{14}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{14}}{2*-1}=\frac{-6+2\sqrt{14}}{-2}
$