Question

How do you solve using the quadratic form? 2x^2+3x-3=-3x-4

Answer 1

Do you mean the quadratic formula?
`x=-0.1771243445 or x=-2.822875656`

Explanation:

`2x^2+3x-3=-3x-4` collect like terms on the LHS

`2x^2+6x+1=0`

`a=2, b=6 and c=1`
`x=[-6\pmsqrt(6^2-[4xx2xx1])]/[2xx2]`

`x=[-6\pmsqrt(36-8)]/[4]`

`x=[-6\pmsqrt28]/[4]`

`x=[-6\pm2sqrt7]/[4]`

`x=(-3+sqrt7)/2` or `x=(-3-sqrt7)/2`

`x=-0.1771243445 or x=-2.822875656`

Answer 2

`2 x^2+3 x-3=-3 x-4`

Explanation:

`a x^2+b x+c=0`
`x_(1,2)=(-b +- \sqrt(b^2-4*a*c))/(2*a)`

Write your equation in the form: `a x^2+b x+c=0`
`2 x^2+3 x-3=-3 x-4` | `+3x`
`2 x^2+3 x-3 +3x=-3 x-4 +3x` |`+4`
`2 x^2+6 x-3+4=-4 +4`
`2 x^2+6 x+1=0`

`a=2`
`b=6`
`c=1`

`x_(1,2)=(-6 +- \sqrt(6^2-4*2*1))/(2*2)=`
`=(-6 +- \sqrt(36-8))/(4)=`
`=(-6 +- \sqrt(28))/(4)=`
`=(-6 +- \sqrt(2^2*7))/(4)=`
`=(-6 +- 2* \sqrt(7))/(4)=`
`=(-3 +- \sqrt(7))/(2)`

`x_1=(-3 + \sqrt(7))/(2)`
`x_2=(-3 - \sqrt(7))/(2)`

Answer 3

`x=(-3+sqrt7)/2 or x=(-3-sqrt7)/2`

Explanation:

Here,

`2x^2+3x-3=-3x-4`

`=>2x^2+3x-3+3x+4=0`

`=>2x^2+6x+1=0`

Comparing with `ax^2+bx+c=0`,

`a=2,b=6 and c=1`

So,

`triangle=b^2-4ac=36-4(2)(1)=36-8=28`

`sqrt(triangle)=2sqrt7`

`:.x=(-b+-sqrt(triangle))/(2a)=(-6+-2sqrt7)/(2(2))=(-3+-sqrt7)/2`

Hence,

`x=(-3+sqrt7)/2 or x=(-3-sqrt7)/2`