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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

17t^2+17t-10=0

a = 17; b = 17; c = -10;
Δ = b2-4ac
Δ = 172-4·17·(-10)
Δ = 969
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{-(17)-\sqrt{969}}{2*17}=\frac{-17-\sqrt{969}}{34}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{969}}{2*17}=\frac{-17+\sqrt{969}}{34}
$