63t^2-29t-4=0

Solution for 63t^2-29t-4=0 equation:

63t^2-29t-4=0

a = 63; b = -29; c = -4;
Δ = b2-4ac
Δ = -292-4·63·(-4)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-43}{2*63}=\frac{-14}{126}
=-1/9
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+43}{2*63}=\frac{72}{126}
=4/7
$