t^2+16t=49

Solution for t^2+16t=49 equation:

t^2+16t=49

We move all terms to the left:

t^2+16t-(49)=0

a = 1; b = 16; c = -49;
Δ = b2-4ac
Δ = 162-4·1·(-49)
Δ = 452
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:

$\sqrt{\Delta}=\sqrt{452}=\sqrt{4*113}=\sqrt{4}*\sqrt{113}=2\sqrt{113}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{113}}{2*1}=\frac{-16-2\sqrt{113}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{113}}{2*1}=\frac{-16+2\sqrt{113}}{2}
$