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

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

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

We move all terms to the left:

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

We multiply parentheses

8x-(7-2(x+1)x)-12=0


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

7-2(x+1)x

determiningTheFunctionDomain
-2(x+1)x+7

We multiply parentheses

-2x^2-2x+7

Back to the equation:

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


We get rid of parentheses

2x^2+2x+8x-7-12=0

We add all the numbers together, and all the variables

2x^2+10x-19=0

a = 2; b = 10; c = -19;
Δ = b2-4ac
Δ = 102-4·2·(-19)
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{7}}{2*2}=\frac{-10-6\sqrt{7}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{7}}{2*2}=\frac{-10+6\sqrt{7}}{4}
$