0.03x(x-4)=0,02(x+6)

Solution for 0.03x(x-4)=0,02(x+6) equation:

0.03x(x-4)=0.02(x+6)

We move all terms to the left:

0.03x(x-4)-(0.02(x+6))=0

We multiply parentheses

0x^2+0x-(0.02(x+6))=0


We calculate terms in parentheses: -(0.02(x+6)), so:

0.02(x+6)

We multiply parentheses

0.02x+0.12

Back to the equation:

-(0.02x+0.12)


We add all the numbers together, and all the variables

x^2+x-(0.02x+0.12)=0

We get rid of parentheses

x^2+x-0.02x-0.12=0

We add all the numbers together, and all the variables

x^2+0.98x-0.12=0

a = 1; b = 0.98; c = -0.12;
Δ = b2-4ac
Δ = 0.982-4·1·(-0.12)
Δ = 1.4404
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.98)-\sqrt{1.4404}}{2*1}=\frac{-0.98-\sqrt{1.4404}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.98)+\sqrt{1.4404}}{2*1}=\frac{-0.98+\sqrt{1.4404}}{2}
$