Y=4x^2-40x+98

Solution for Y=4x^2-40x+98 equation:

=4Y^2-40Y+98

We move all terms to the left:

-(4Y^2-40Y+98)=0

We get rid of parentheses

-4Y^2+40Y-98=0

a = -4; b = 40; c = -98;
Δ = b2-4ac
Δ = 402-4·(-4)·(-98)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{2}}{2*-4}=\frac{-40-4\sqrt{2}}{-8}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{2}}{2*-4}=\frac{-40+4\sqrt{2}}{-8}
$