0.11x+0.7x(x-2)=0.01(4x-3)

Solution for 0.11x+0.7x(x-2)=0.01(4x-3) equation:

0.11x+0.7x(x-2)=0.01(4x-3)

We move all terms to the left:

0.11x+0.7x(x-2)-(0.01(4x-3))=0

We multiply parentheses

0x^2+0.11x+0x-(0.01(4x-3))=0


We calculate terms in parentheses: -(0.01(4x-3)), so:

0.01(4x-3)

We multiply parentheses

0.04x-0.03

Back to the equation:

-(0.04x-0.03)


We add all the numbers together, and all the variables

x^2+1.11x-(0.04x-0.03)=0

We get rid of parentheses

x^2+1.11x-0.04x+0.03=0

We add all the numbers together, and all the variables

x^2+1.07x+0.03=0

a = 1; b = 1.07; c = +0.03;
Δ = b2-4ac
Δ = 1.072-4·1·0.03
Δ = 1.0249
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{-(1.07)-\sqrt{1.0249}}{2*1}=\frac{-1.07-\sqrt{1.0249}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.07)+\sqrt{1.0249}}{2*1}=\frac{-1.07+\sqrt{1.0249}}{2}
$