10t2+-5=t+150

Solution for 10t2+-5=t+150 equation:

10t^2+-5=t+150

We move all terms to the left:

10t^2+-5-(t+150)=0

We add all the numbers together, and all the variables

10t^2-(t+150)=0

We get rid of parentheses

10t^2-t-150=0

We add all the numbers together, and all the variables

10t^2-1t-150=0

a = 10; b = -1; c = -150;
Δ = b2-4ac
Δ = -12-4·10·(-150)
Δ = 6001
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{6001}}{2*10}=\frac{1-\sqrt{6001}}{20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{6001}}{2*10}=\frac{1+\sqrt{6001}}{20}
$