4t^2+10t=16

Solution for 4t^2+10t=16 equation:

4t^2+10t=16

We move all terms to the left:

4t^2+10t-(16)=0

a = 4; b = 10; c = -16;
Δ = b2-4ac
Δ = 102-4·4·(-16)
Δ = 356
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:

$\sqrt{\Delta}=\sqrt{356}=\sqrt{4*89}=\sqrt{4}*\sqrt{89}=2\sqrt{89}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{89}}{2*4}=\frac{-10-2\sqrt{89}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{89}}{2*4}=\frac{-10+2\sqrt{89}}{8}
$