10t2+-5=t+15

Solution for 10t2+-5=t+15 equation:

10t^2+-5=t+15

We move all terms to the left:

10t^2+-5-(t+15)=0

We add all the numbers together, and all the variables

10t^2-(t+15)=0

We get rid of parentheses

10t^2-t-15=0

We add all the numbers together, and all the variables

10t^2-1t-15=0

a = 10; b = -1; c = -15;
Δ = b2-4ac
Δ = -12-4·10·(-15)
Δ = 601
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{-(-1)-\sqrt{601}}{2*10}=\frac{1-\sqrt{601}}{20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{601}}{2*10}=\frac{1+\sqrt{601}}{20}
$