10-t^2+t*2=0

Solution for 10-t^2+t*2=0 equation:

10-t^2+t*2=0

We add all the numbers together, and all the variables

-1t^2+t*2+10=0

Wy multiply elements

-1t^2+2t+10=0

a = -1; b = 2; c = +10;
Δ = b2-4ac
Δ = 22-4·(-1)·10
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*-1}=\frac{-2-2\sqrt{11}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{11}}{2*-1}=\frac{-2+2\sqrt{11}}{-2}
$