-2t^2+5t-8=-t-5

Solution for -2t^2+5t-8=-t-5 equation:

-2t^2+5t-8=-t-5

We move all terms to the left:

-2t^2+5t-8-(-t-5)=0

We add all the numbers together, and all the variables

-2t^2+5t-(-1t-5)-8=0

We get rid of parentheses

-2t^2+5t+1t+5-8=0

We add all the numbers together, and all the variables

-2t^2+6t-3=0

a = -2; b = 6; c = -3;
Δ = b2-4ac
Δ = 62-4·(-2)·(-3)
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{3}}{2*-2}=\frac{-6-2\sqrt{3}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{3}}{2*-2}=\frac{-6+2\sqrt{3}}{-4}
$