(4t+3t^2)=+18(9t+2)

Solution for (4t+3t^2)=+18(9t+2) equation:

(4t+3t^2)=+18(9t+2)

We move all terms to the left:

(4t+3t^2)-(+18(9t+2))=0

We get rid of parentheses

3t^2+4t-(+18(9t+2))=0


We calculate terms in parentheses: -(+18(9t+2)), so:

+18(9t+2)

determiningTheFunctionDomain
18(9t+2)

We multiply parentheses

162t+36

Back to the equation:

-(162t+36)


We get rid of parentheses

3t^2+4t-162t-36=0

We add all the numbers together, and all the variables

3t^2-158t-36=0

a = 3; b = -158; c = -36;
Δ = b2-4ac
Δ = -1582-4·3·(-36)
Δ = 25396
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:

$\sqrt{\Delta}=\sqrt{25396}=\sqrt{4*6349}=\sqrt{4}*\sqrt{6349}=2\sqrt{6349}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-158)-2\sqrt{6349}}{2*3}=\frac{158-2\sqrt{6349}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-158)+2\sqrt{6349}}{2*3}=\frac{158+2\sqrt{6349}}{6}
$