(3x+2)=(4x^2-7x+1)

Solution for (3x+2)=(4x^2-7x+1) equation:

(3x+2)=(4x^2-7x+1)

We move all terms to the left:

(3x+2)-((4x^2-7x+1))=0

We get rid of parentheses

3x-((4x^2-7x+1))+2=0


We calculate terms in parentheses: -((4x^2-7x+1)), so:

(4x^2-7x+1)

We get rid of parentheses

4x^2-7x+1

Back to the equation:

-(4x^2-7x+1)


We get rid of parentheses

-4x^2+3x+7x-1+2=0

We add all the numbers together, and all the variables

-4x^2+10x+1=0

a = -4; b = 10; c = +1;
Δ = b2-4ac
Δ = 102-4·(-4)·1
Δ = 116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{29}}{2*-4}=\frac{-10-2\sqrt{29}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{29}}{2*-4}=\frac{-10+2\sqrt{29}}{-8}
$