4(t-3)+8=4(t2t-5)

Solution for 4(t-3)+8=4(t2t-5) equation:

4(t-3)+8=4(t2t-5)

We move all terms to the left:

4(t-3)+8-(4(t2t-5))=0

We multiply parentheses

4t-(4(t2t-5))-12+8=0


We calculate terms in parentheses: -(4(t2t-5)), so:

4(t2t-5)

We multiply parentheses

4t^2-20

Back to the equation:

-(4t^2-20)


We add all the numbers together, and all the variables

4t-(4t^2-20)-4=0

We get rid of parentheses

-4t^2+4t+20-4=0

We add all the numbers together, and all the variables

-4t^2+4t+16=0

a = -4; b = 4; c = +16;
Δ = b2-4ac
Δ = 42-4·(-4)·16
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{17}}{2*-4}=\frac{-4-4\sqrt{17}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{17}}{2*-4}=\frac{-4+4\sqrt{17}}{-8}
$