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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

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

We multiply parentheses

8t-3t+6+12

We add all the numbers together, and all the variables

5t+18

Back to the equation:

-(5t+18)


We get rid of parentheses

4t^2-2t+10t+t-5t-5-18=0

We add all the numbers together, and all the variables

4t^2+4t-23=0

a = 4; b = 4; c = -23;
Δ = b2-4ac
Δ = 42-4·4·(-23)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{6}}{2*4}=\frac{-4-8\sqrt{6}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{6}}{2*4}=\frac{-4+8\sqrt{6}}{8}
$