0=(25-3x)(x+16)

Solution for 0=(25-3x)(x+16) equation:

0=(25-3x)(x+16)

We move all terms to the left:

0-((25-3x)(x+16))=0

We add all the numbers together, and all the variables

-((-3x+25)(x+16))+0=0

We add all the numbers together, and all the variables

-((-3x+25)(x+16))=0

We multiply parentheses ..

-((-3x^2-48x+25x+400))=0


We calculate terms in parentheses: -((-3x^2-48x+25x+400)), so:

(-3x^2-48x+25x+400)

We get rid of parentheses

-3x^2-48x+25x+400

We add all the numbers together, and all the variables

-3x^2-23x+400

Back to the equation:

-(-3x^2-23x+400)


We get rid of parentheses

3x^2+23x-400=0

a = 3; b = 23; c = -400;
Δ = b2-4ac
Δ = 232-4·3·(-400)
Δ = 5329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5329}=73$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-73}{2*3}=\frac{-96}{6}
=-16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+73}{2*3}=\frac{50}{6}
=8+1/3
$