X=-t2+3t+10

Solution for X=-t2+3t+10 equation:

X=-X2+3X+10

We move all terms to the left:

X-(-X2+3X+10)=0

We add all the numbers together, and all the variables

-(-1X^2+3X+10)+X=0

We get rid of parentheses

1X^2-3X+X-10=0

We add all the numbers together, and all the variables

X^2-2X-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:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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