t-(2t+5)(1-2t)=2(3+4t)-3(t-4)t=

Solution for t-(2t+5)(1-2t)=2(3+4t)-3(t-4)t= equation:

t-(2t+5)(1-2t)=2(3+4t)-3(t-4)t=

We move all terms to the left:

t-(2t+5)(1-2t)-(2(3+4t)-3(t-4)t)=0

We add all the numbers together, and all the variables

t-(2t+5)(-2t+1)-(2(4t+3)-3(t-4)t)=0

We multiply parentheses ..

-(-4t^2+2t-10t+5)+t-(2(4t+3)-3(t-4)t)=0


We calculate terms in parentheses: -(2(4t+3)-3(t-4)t), so:

2(4t+3)-3(t-4)t

We multiply parentheses

-3t^2+8t+12t+6

We add all the numbers together, and all the variables

-3t^2+20t+6

Back to the equation:

-(-3t^2+20t+6)


We get rid of parentheses

4t^2+3t^2-2t+10t-20t+t-5-6=0

We add all the numbers together, and all the variables

7t^2-11t-11=0

a = 7; b = -11; c = -11;
Δ = b2-4ac
Δ = -112-4·7·(-11)
Δ = 429
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{-(-11)-\sqrt{429}}{2*7}=\frac{11-\sqrt{429}}{14}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{429}}{2*7}=\frac{11+\sqrt{429}}{14}
$