-3t(t-5)+6t=5t-1

Solution for -3t(t-5)+6t=5t-1 equation:

-3t(t-5)+6t=5t-1

We move all terms to the left:

-3t(t-5)+6t-(5t-1)=0

We add all the numbers together, and all the variables

6t-3t(t-5)-(5t-1)=0

We multiply parentheses

-3t^2+6t+15t-(5t-1)=0

We get rid of parentheses

-3t^2+6t+15t-5t+1=0

We add all the numbers together, and all the variables

-3t^2+16t+1=0

a = -3; b = 16; c = +1;
Δ = b2-4ac
Δ = 162-4·(-3)·1
Δ = 268
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:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{67}}{2*-3}=\frac{-16-2\sqrt{67}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{67}}{2*-3}=\frac{-16+2\sqrt{67}}{-6}
$