(1=4x)(7x-5)=0

Solution for (1=4x)(7x-5)=0 equation:

(1=4x)(7x-5)=0

We move all terms to the left:

(1-(4x)(7x-5))=0


We calculate terms in parentheses: +(1-4x(7x-5)), so:

1-4x(7x-5)

determiningTheFunctionDomain
-4x(7x-5)+1

We multiply parentheses

-28x^2+20x+1

Back to the equation:

+(-28x^2+20x+1)


We get rid of parentheses

-28x^2+20x+1=0

a = -28; b = 20; c = +1;
Δ = b2-4ac
Δ = 202-4·(-28)·1
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16\sqrt{2}}{2*-28}=\frac{-20-16\sqrt{2}}{-56}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16\sqrt{2}}{2*-28}=\frac{-20+16\sqrt{2}}{-56}
$