(3t+5)(t-3)+(3t+5)(2t+7)=0

Solution for (3t+5)(t-3)+(3t+5)(2t+7)=0 equation:

(3t+5)(t-3)+(3t+5)(2t+7)=0

We multiply parentheses ..

(+3t^2-9t+5t-15)+(3t+5)(2t+7)=0

We get rid of parentheses

3t^2-9t+5t+(3t+5)(2t+7)-15=0

We multiply parentheses ..

3t^2+(+6t^2+21t+10t+35)-9t+5t-15=0

We add all the numbers together, and all the variables

3t^2+(+6t^2+21t+10t+35)-4t-15=0

We get rid of parentheses

3t^2+6t^2+21t+10t-4t+35-15=0

We add all the numbers together, and all the variables

9t^2+27t+20=0

a = 9; b = 27; c = +20;
Δ = b2-4ac
Δ = 272-4·9·20
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{-(27)-3}{2*9}=\frac{-30}{18}
=-1+2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+3}{2*9}=\frac{-24}{18}
=-1+1/3
$