7t^2-42t=14=0

Solution for 7t^2-42t=14=0 equation:

7t^2-42t=14=0

We move all terms to the left:

7t^2-42t-(14)=0

a = 7; b = -42; c = -14;
Δ = b2-4ac
Δ = -422-4·7·(-14)
Δ = 2156
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:

$\sqrt{\Delta}=\sqrt{2156}=\sqrt{196*11}=\sqrt{196}*\sqrt{11}=14\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-14\sqrt{11}}{2*7}=\frac{42-14\sqrt{11}}{14}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+14\sqrt{11}}{2*7}=\frac{42+14\sqrt{11}}{14}
$