4t^2=29t-7

Solution for 4t^2=29t-7 equation:

4t^2=29t-7

We move all terms to the left:

4t^2-(29t-7)=0

We get rid of parentheses

4t^2-29t+7=0

a = 4; b = -29; c = +7;
Δ = b2-4ac
Δ = -292-4·4·7
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-27}{2*4}=\frac{2}{8}
=1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+27}{2*4}=\frac{56}{8}
=7
$