2t^2+t=28

Solution for 2t^2+t=28 equation:

2t^2+t=28

We move all terms to the left:

2t^2+t-(28)=0

a = 2; b = 1; c = -28;
Δ = b2-4ac
Δ = 12-4·2·(-28)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-15}{2*2}=\frac{-16}{4}
=-4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+15}{2*2}=\frac{14}{4}
=3+1/2
$