A=(3t^2+7t+4)

Solution for A=(3t^2+7t+4) equation:

=(3A^2+7A+4)

We move all terms to the left:

-((3A^2+7A+4))=0


We calculate terms in parentheses: -((3A^2+7A+4)), so:

(3A^2+7A+4)

We get rid of parentheses

3A^2+7A+4

Back to the equation:

-(3A^2+7A+4)


We get rid of parentheses

-3A^2-7A-4=0

a = -3; b = -7; c = -4;
Δ = b2-4ac
Δ = -72-4·(-3)·(-4)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*-3}=\frac{6}{-6}
=-1
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*-3}=\frac{8}{-6}
=-1+1/3
$