t+2(t-4)=5(1-t2)

Solution for t+2(t-4)=5(1-t2) equation:

t+2(t-4)=5(1-t2)

We move all terms to the left:

t+2(t-4)-(5(1-t2))=0

We add all the numbers together, and all the variables

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

We multiply parentheses

-(5(-1t^2+1))+t+2t-8=0


We calculate terms in parentheses: -(5(-1t^2+1)), so:

5(-1t^2+1)

We multiply parentheses

-5t^2+5

Back to the equation:

-(-5t^2+5)


We add all the numbers together, and all the variables

-(-5t^2+5)+3t-8=0

We get rid of parentheses

5t^2+3t-5-8=0

We add all the numbers together, and all the variables

5t^2+3t-13=0

a = 5; b = 3; c = -13;
Δ = b2-4ac
Δ = 32-4·5·(-13)
Δ = 269
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{-(3)-\sqrt{269}}{2*5}=\frac{-3-\sqrt{269}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{269}}{2*5}=\frac{-3+\sqrt{269}}{10}
$