-9(t+2)=4(t-15)t=

Solution for -9(t+2)=4(t-15)t= equation:

-9(t+2)=4(t-15)t=

We move all terms to the left:

-9(t+2)-(4(t-15)t)=0

We multiply parentheses

-9t-(4(t-15)t)-18=0


We calculate terms in parentheses: -(4(t-15)t), so:

4(t-15)t

We multiply parentheses

4t^2-60t

Back to the equation:

-(4t^2-60t)


We get rid of parentheses

-4t^2-9t+60t-18=0

We add all the numbers together, and all the variables

-4t^2+51t-18=0

a = -4; b = 51; c = -18;
Δ = b2-4ac
Δ = 512-4·(-4)·(-18)
Δ = 2313
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:

$\sqrt{\Delta}=\sqrt{2313}=\sqrt{9*257}=\sqrt{9}*\sqrt{257}=3\sqrt{257}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-3\sqrt{257}}{2*-4}=\frac{-51-3\sqrt{257}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+3\sqrt{257}}{2*-4}=\frac{-51+3\sqrt{257}}{-8}
$