0=(23t^2-29t+6)

Solution for 0=(23t^2-29t+6) equation:

0=(23t^2-29t+6)

We move all terms to the left:

0-((23t^2-29t+6))=0

We add all the numbers together, and all the variables

-((23t^2-29t+6))=0


We calculate terms in parentheses: -((23t^2-29t+6)), so:

(23t^2-29t+6)

We get rid of parentheses

23t^2-29t+6

Back to the equation:

-(23t^2-29t+6)


We get rid of parentheses

-23t^2+29t-6=0

a = -23; b = 29; c = -6;
Δ = b2-4ac
Δ = 292-4·(-23)·(-6)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-17}{2*-23}=\frac{-46}{-46}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+17}{2*-23}=\frac{-12}{-46}
=6/23
$