(2t+1)(t-4)=t^2-4t-6

Solution for (2t+1)(t-4)=t^2-4t-6 equation:

(2t+1)(t-4)=t^2-4t-6

We move all terms to the left:

(2t+1)(t-4)-(t^2-4t-6)=0

We get rid of parentheses

-t^2+(2t+1)(t-4)+4t+6=0

We multiply parentheses ..

-t^2+(+2t^2-8t+t-4)+4t+6=0

We add all the numbers together, and all the variables

-1t^2+(+2t^2-8t+t-4)+4t+6=0

We get rid of parentheses

-1t^2+2t^2-8t+t+4t-4+6=0

We add all the numbers together, and all the variables

t^2-3t+2=0

a = 1; b = -3; c = +2;
Δ = b2-4ac
Δ = -32-4·1·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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