t2+14t=49

Solution for t2+14t=49 equation:

t2+14t=49

We move all terms to the left:

t2+14t-(49)=0

We add all the numbers together, and all the variables

t^2+14t-49=0

a = 1; b = 14; c = -49;
Δ = b2-4ac
Δ = 142-4·1·(-49)
Δ = 392
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:

$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14\sqrt{2}}{2*1}=\frac{-14-14\sqrt{2}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14\sqrt{2}}{2*1}=\frac{-14+14\sqrt{2}}{2}
$