(2t+1)(t-4)=X

Solution for (2t+1)(t-4)=X equation:

(2t+1)(t-4)=t

We move all terms to the left:

(2t+1)(t-4)-(t)=0

We add all the numbers together, and all the variables

-1t+(2t+1)(t-4)=0

We multiply parentheses ..

(+2t^2-8t+t-4)-1t=0

We get rid of parentheses

2t^2-8t+t-1t-4=0

We add all the numbers together, and all the variables

2t^2-8t-4=0

a = 2; b = -8; c = -4;
Δ = b2-4ac
Δ = -82-4·2·(-4)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{6}}{2*2}=\frac{8-4\sqrt{6}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{6}}{2*2}=\frac{8+4\sqrt{6}}{4}
$