(9t^2+4t-6)=(t^2-2t+4)

Solution for (9t^2+4t-6)=(t^2-2t+4) equation:

(9t^2+4t-6)=(t^2-2t+4)

We move all terms to the left:

(9t^2+4t-6)-((t^2-2t+4))=0

We get rid of parentheses

9t^2+4t-((t^2-2t+4))-6=0


We calculate terms in parentheses: -((t^2-2t+4)), so:

(t^2-2t+4)

We get rid of parentheses

t^2-2t+4

Back to the equation:

-(t^2-2t+4)


We get rid of parentheses

9t^2-t^2+4t+2t-4-6=0

We add all the numbers together, and all the variables

8t^2+6t-10=0

a = 8; b = 6; c = -10;
Δ = b2-4ac
Δ = 62-4·8·(-10)
Δ = 356
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:

$\sqrt{\Delta}=\sqrt{356}=\sqrt{4*89}=\sqrt{4}*\sqrt{89}=2\sqrt{89}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{89}}{2*8}=\frac{-6-2\sqrt{89}}{16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{89}}{2*8}=\frac{-6+2\sqrt{89}}{16}
$