(2t-3)2=2t2+5t-26

Solution for (2t-3)2=2t2+5t-26 equation:

(2t-3)2=2t^2+5t-26

We move all terms to the left:

(2t-3)2-(2t^2+5t-26)=0

We multiply parentheses

4t-(2t^2+5t-26)-6=0

We get rid of parentheses

-2t^2+4t-5t+26-6=0

We add all the numbers together, and all the variables

-2t^2-1t+20=0

a = -2; b = -1; c = +20;
Δ = b2-4ac
Δ = -12-4·(-2)·20
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{161}}{2*-2}=\frac{1-\sqrt{161}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{161}}{2*-2}=\frac{1+\sqrt{161}}{-4}
$