t(t-3)+8=4(2t-5)

Solution for t(t-3)+8=4(2t-5) equation:

t(t-3)+8=4(2t-5)

We move all terms to the left:

t(t-3)+8-(4(2t-5))=0

We multiply parentheses

t^2-3t-(4(2t-5))+8=0


We calculate terms in parentheses: -(4(2t-5)), so:

4(2t-5)

We multiply parentheses

8t-20

Back to the equation:

-(8t-20)


We get rid of parentheses

t^2-3t-8t+20+8=0

We add all the numbers together, and all the variables

t^2-11t+28=0

a = 1; b = -11; c = +28;
Δ = b2-4ac
Δ = -112-4·1·28
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3}{2*1}=\frac{8}{2}
=4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3}{2*1}=\frac{14}{2}
=7
$