(3a^2-7a+6)=(4a^2-3a+4)

Solution for (3a^2-7a+6)=(4a^2-3a+4) equation:

(3a^2-7a+6)=(4a^2-3a+4)

We move all terms to the left:

(3a^2-7a+6)-((4a^2-3a+4))=0

We get rid of parentheses

3a^2-7a-((4a^2-3a+4))+6=0


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

(4a^2-3a+4)

We get rid of parentheses

4a^2-3a+4

Back to the equation:

-(4a^2-3a+4)


We get rid of parentheses

3a^2-4a^2-7a+3a-4+6=0

We add all the numbers together, and all the variables

-1a^2-4a+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:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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