16t^2-29t+7=17

Solution for 16t^2-29t+7=17 equation:

16t^2-29t+7=17

We move all terms to the left:

16t^2-29t+7-(17)=0

We add all the numbers together, and all the variables

16t^2-29t-10=0

a = 16; b = -29; c = -10;
Δ = b2-4ac
Δ = -292-4·16·(-10)
Δ = 1481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-\sqrt{1481}}{2*16}=\frac{29-\sqrt{1481}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{1481}}{2*16}=\frac{29+\sqrt{1481}}{32}
$