Y=-x^2/10+2x+2

Solution for Y=-x^2/10+2x+2 equation:

=-Y^2/10+2Y+2

We move all terms to the left:

-(-Y^2/10+2Y+2)=0


Domain of the equation: 10+2Y+2)!=0

We move all terms containing Y to the left, all other terms to the right

2Y+2)!=-10

Y∈R


We get rid of parentheses

Y^2/10-2Y-2=0

We multiply all the terms by the denominator


Y^2-2Y*10-2*10=0

We add all the numbers together, and all the variables

Y^2-2Y*10-20=0

Wy multiply elements

Y^2-20Y-20=0

a = 1; b = -20; c = -20;
Δ = b2-4ac
Δ = -202-4·1·(-20)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{30}}{2*1}=\frac{20-4\sqrt{30}}{2}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{30}}{2*1}=\frac{20+4\sqrt{30}}{2}
$