(t+24)=t(t+6)

Solution for (t+24)=t(t+6) equation:

(t+24)=t(t+6)

We move all terms to the left:

(t+24)-(t(t+6))=0

We get rid of parentheses

t-(t(t+6))+24=0


We calculate terms in parentheses: -(t(t+6)), so:

t(t+6)

We multiply parentheses

t^2+6t

Back to the equation:

-(t^2+6t)


We get rid of parentheses

-t^2+t-6t+24=0

We add all the numbers together, and all the variables

-1t^2-5t+24=0

a = -1; b = -5; c = +24;
Δ = b2-4ac
Δ = -52-4·(-1)·24
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*-1}=\frac{-6}{-2}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*-1}=\frac{16}{-2}
=-8
$