(T+2)=5x^2-2x

Solution for (T+2)=5x^2-2x equation:

(+2)=5T^2-2T

We move all terms to the left:

(+2)-(5T^2-2T)=0

We add all the numbers together, and all the variables

-(5T^2-2T)+2=0

We get rid of parentheses

-5T^2+2T+2=0

a = -5; b = 2; c = +2;
Δ = b2-4ac
Δ = 22-4·(-5)·2
Δ = 44
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{11}}{2*-5}=\frac{-2-2\sqrt{11}}{-10}
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{11}}{2*-5}=\frac{-2+2\sqrt{11}}{-10}
$