9x-3/7=5x^2+8x-4

Solution for 9x-3/7=5x^2+8x-4 equation:

9x-3/7=5x^2+8x-4

We move all terms to the left:

9x-3/7-(5x^2+8x-4)=0

We get rid of parentheses

-5x^2+9x-8x+4-3/7=0

We multiply all the terms by the denominator


-5x^2*7+9x*7-8x*7-3+4*7=0

We add all the numbers together, and all the variables

-5x^2*7+9x*7-8x*7+25=0

Wy multiply elements

-35x^2+63x-56x+25=0

We add all the numbers together, and all the variables

-35x^2+7x+25=0

a = -35; b = 7; c = +25;
Δ = b2-4ac
Δ = 72-4·(-35)·25
Δ = 3549
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:

$\sqrt{\Delta}=\sqrt{3549}=\sqrt{169*21}=\sqrt{169}*\sqrt{21}=13\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13\sqrt{21}}{2*-35}=\frac{-7-13\sqrt{21}}{-70}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13\sqrt{21}}{2*-35}=\frac{-7+13\sqrt{21}}{-70}
$