t^2*28=3t

Solution for t^2*28=3t equation:

t^2*28=3t

We move all terms to the left:

t^2*28-(3t)=0

Wy multiply elements

28t^2-3t=0

a = 28; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·28·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*28}=\frac{0}{56}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*28}=\frac{6}{56}
=3/28
$