0,7^2x*0,7^2-5x-0,4^1-3x=-330/49

Solution for 0,7^2x*0,7^2-5x-0,4^1-3x=-330/49 equation:

0.7^2x*0.7^2-5x-0.4^1-3x=-330/49

We move all terms to the left:

0.7^2x*0.7^2-5x-0.4^1-3x-(-330/49)=0

We add all the numbers together, and all the variables

-8x+0.7^2x*0.7^2-0.4-(-330/49)=0

We get rid of parentheses

-8x+0.7^2x*0.7^2-0.4+330/49=0

We multiply all the terms by the denominator


-8x*49+(0.7^2x*0.7^2)*49+330-(0.4)*49=0

We add all the numbers together, and all the variables

-8x*49+(0.7^2x*0.7^2)*49+310.4=0

We multiply parentheses

0x^2-8x*49+310.4=0

Wy multiply elements

0x^2-392x+310.4=0

We add all the numbers together, and all the variables

x^2-392x+310.4=0

a = 1; b = -392; c = +310.4;
Δ = b2-4ac
Δ = -3922-4·1·310.4
Δ = 152422.4
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{-(-392)-\sqrt{152422.4}}{2*1}=\frac{392-\sqrt{152422.4}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-392)+\sqrt{152422.4}}{2*1}=\frac{392+\sqrt{152422.4}}{2}
$