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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

10t^2+t-2t-5t-(2(4t+3)-3(t-4))-5=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 add all the numbers together, and all the variables

10t^2-6t-(5t+18)-5=0

We get rid of parentheses

10t^2-6t-5t-18-5=0

We add all the numbers together, and all the variables

10t^2-11t-23=0

a = 10; b = -11; c = -23;
Δ = b2-4ac
Δ = -112-4·10·(-23)
Δ = 1041
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{1041}}{2*10}=\frac{11-\sqrt{1041}}{20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{1041}}{2*10}=\frac{11+\sqrt{1041}}{20}
$