(Y-2)^2+(y-4)(y-7)=4

Solution for (Y-2)^2+(y-4)(y-7)=4 equation:

(-2)^2+(Y-4)(Y-7)=4

We move all terms to the left:

(-2)^2+(Y-4)(Y-7)-(4)=0

determiningTheFunctionDomain
(Y-4)(Y-7)-4+(-2)^2=0

We add all the numbers together, and all the variables

(Y-4)(Y-7)=0

We multiply parentheses ..

(+Y^2-7Y-4Y+28)=0

We get rid of parentheses

Y^2-7Y-4Y+28=0

We add all the numbers together, and all the variables

Y^2-11Y+28=0

a = 1; b = -11; c = +28;
Δ = b2-4ac
Δ = -112-4·1·28
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3}{2*1}=\frac{8}{2}
=4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3}{2*1}=\frac{14}{2}
=7
$