3.5(x-2)+40=-0.3x^2+160

Solution for 3.5(x-2)+40=-0.3x^2+160 equation:

3.5(x-2)+40=-0.3x^2+160

We move all terms to the left:

3.5(x-2)+40-(-0.3x^2+160)=0

We multiply parentheses

-(-0.3x^2+160)+3.5x-7+40=0

We get rid of parentheses

0.3x^2+3.5x-160-7+40=0

We add all the numbers together, and all the variables

0.3x^2+3.5x-127=0

a = 0.3; b = 3.5; c = -127;
Δ = b2-4ac
Δ = 3.52-4·0.3·(-127)
Δ = 164.65
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{-(3.5)-\sqrt{164.65}}{2*0.3}=\frac{-3.5-\sqrt{164.65}}{0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.5)+\sqrt{164.65}}{2*0.3}=\frac{-3.5+\sqrt{164.65}}{0.6}
$