3t+(t-15)4+2t(2)=435

Solution for 3t+(t-15)4+2t(2)=435 equation:

3t+(t-15)4+2t(2)=435

We move all terms to the left:

3t+(t-15)4+2t(2)-(435)=0

We add all the numbers together, and all the variables

2t^2+3t+(t-15)4-435=0

We multiply parentheses

2t^2+3t+4t-60-435=0

We add all the numbers together, and all the variables

2t^2+7t-495=0

a = 2; b = 7; c = -495;
Δ = b2-4ac
Δ = 72-4·2·(-495)
Δ = 4009
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{4009}}{2*2}=\frac{-7-\sqrt{4009}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{4009}}{2*2}=\frac{-7+\sqrt{4009}}{4}
$