17t^2+20t+3=2t+13

Solution for 17t^2+20t+3=2t+13 equation:

17t^2+20t+3=2t+13

We move all terms to the left:

17t^2+20t+3-(2t+13)=0

We get rid of parentheses

17t^2+20t-2t-13+3=0

We add all the numbers together, and all the variables

17t^2+18t-10=0

a = 17; b = 18; c = -10;
Δ = b2-4ac
Δ = 182-4·17·(-10)
Δ = 1004
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:

$\sqrt{\Delta}=\sqrt{1004}=\sqrt{4*251}=\sqrt{4}*\sqrt{251}=2\sqrt{251}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{251}}{2*17}=\frac{-18-2\sqrt{251}}{34}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{251}}{2*17}=\frac{-18+2\sqrt{251}}{34}
$