(t)=(t^2+5t-2)

Solution for (t)=(t^2+5t-2) equation:

(t)=(t^2+5t-2)

We move all terms to the left:

(t)-((t^2+5t-2))=0


We calculate terms in parentheses: -((t^2+5t-2)), so:

(t^2+5t-2)

We get rid of parentheses

t^2+5t-2

Back to the equation:

-(t^2+5t-2)


We get rid of parentheses

-t^2+t-5t+2=0

We add all the numbers together, and all the variables

-1t^2-4t+2=0

a = -1; b = -4; c = +2;
Δ = b2-4ac
Δ = -42-4·(-1)·2
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{6}}{2*-1}=\frac{4-2\sqrt{6}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{6}}{2*-1}=\frac{4+2\sqrt{6}}{-2}
$