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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

2t^2+5t+10t-(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

2t^2+15t-(5t+18)-5=0

We get rid of parentheses

2t^2+15t-5t-18-5=0

We add all the numbers together, and all the variables

2t^2+10t-23=0

a = 2; b = 10; c = -23;
Δ = b2-4ac
Δ = 102-4·2·(-23)
Δ = 284
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:

$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{71}}{2*2}=\frac{-10-2\sqrt{71}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{71}}{2*2}=\frac{-10+2\sqrt{71}}{4}
$