x(7x+4)=2x^2-4x+21

Solution for x(7x+4)=2x^2-4x+21 equation:

x(7x+4)=2x^2-4x+21

We move all terms to the left:

x(7x+4)-(2x^2-4x+21)=0

We multiply parentheses

7x^2+4x-(2x^2-4x+21)=0

We get rid of parentheses

7x^2-2x^2+4x+4x-21=0

We add all the numbers together, and all the variables

5x^2+8x-21=0

a = 5; b = 8; c = -21;
Δ = b2-4ac
Δ = 82-4·5·(-21)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-22}{2*5}=\frac{-30}{10}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+22}{2*5}=\frac{14}{10}
=1+2/5
$