(4x^2+15x+150)=(3x^2-15x+25)

Solution for (4x^2+15x+150)=(3x^2-15x+25) equation:

(4x^2+15x+150)=(3x^2-15x+25)

We move all terms to the left:

(4x^2+15x+150)-((3x^2-15x+25))=0

We get rid of parentheses

4x^2+15x-((3x^2-15x+25))+150=0


We calculate terms in parentheses: -((3x^2-15x+25)), so:

(3x^2-15x+25)

We get rid of parentheses

3x^2-15x+25

Back to the equation:

-(3x^2-15x+25)


We get rid of parentheses

4x^2-3x^2+15x+15x-25+150=0

We add all the numbers together, and all the variables

x^2+30x+125=0

a = 1; b = 30; c = +125;
Δ = b2-4ac
Δ = 302-4·1·125
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-20}{2*1}=\frac{-50}{2}
=-25
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+20}{2*1}=\frac{-10}{2}
=-5
$