3t^2=-t+10

Solution for 3t^2=-t+10 equation:

3t^2=-t+10

We move all terms to the left:

3t^2-(-t+10)=0

We add all the numbers together, and all the variables

3t^2-(-1t+10)=0

We get rid of parentheses

3t^2+1t-10=0

We add all the numbers together, and all the variables

3t^2+t-10=0

a = 3; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·3·(-10)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*3}=\frac{-12}{6}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*3}=\frac{10}{6}
=1+2/3
$