# Solve this equation using the quadratic formula: 2x(x - 4) - 3(x + 5) = x(1 - x) - 16 Enter two simplified solutions

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Solve `2x(x - 4) - 3(x + 5) = x(1 - x) - 16` using the quadratic formula.

First simplify the expression.

`2x(x - 4) - 3(x + 5) = x(1 - x) - 16`

`2x^2 - 8x - 3x - 15 = x - x^2 - 16`

`3x^2 - 12x + 1 = 0`

Quadratic formula is: `(-b+-sqrt(b^2 - 4ac))/(2a)`

For this equation, a = 3, b = -12, and c = 1.

`(12+-sqrt(12^2 - 4(3)(1)))/(2*3)`

`(12+-sqrt(144-12))/6`

`(12 +-sqrt(132))/6` = `(12 +-2sqrt(33))/6` Cancel every term by 2.

`(6 +-sqrt(33))/3`

**The solution using the quadratic formula is `(6 +-sqrt(33))/3` or `x = 3.915, 0.085` **

To use the quadratic formula, we need to get the function into the form `ax^2 + bx + c = 0` We do that by completing the calculations of the original function.

`2x(x-4)-3(x+5)=x(1-x)-16`

`2x^2-8x-3x-15=x-x^2-16` combine like terms

`2x^2-11x-15=-x^2+x-16 `

`3x^2-10x+1=0 `

`b=-10, a=3 c=1 x=(-b+-sqrt(b^2-4ac))/(2a)`

`(10+-sqrt(-10^2-4(3)(1)))/(2 xx 3)=(10+-sqrt(100-12))/6=`

`(10+-sqrt88)/6=10+-2sqrt(22)/6=(5+-sqrt22)/3`

`x~~3.23, 0.10`

2x(x - 4) - 3(x + 5) = x(1 - x) - 16

distribute the numbers outside of the parentheses

2x^2-8x-3x-15=x-x^2-16

combine like terms

2x^2-11x-15=-x^2+x-16

get the terms on the same side

2x^2-11x-15+x^2-x+16=0

combine like terms

3x^2-12x+1=0

a=3 b=-12 c=1

the quadratic formula is: `(-b+-sqrt(b^2-4ac)) / (2a)`

plug in the numbers

`(12+-sqrt(12^2-4(3)(1)))/((2)(3))`

`(12+-sqrt(144-12))/(6) `

`(12+-sqrt(132))/(6) `

`sqrt(132)=11.6619038`

`(12+-11.6619038) / 6 `

`(12+11.6619038) / 6 = (23.6619038) / 6 =3.94365063333 or 3.94`

`(12-11.6619038) / 6 = (0.3380962) / 6 =0.05634936666 or .056`

**x= 3.94 x= .056**