Question

How do you solve this quadratic equation `2x^2- 10+ 7=0`?

Answer 1

`x=+-1/2sqrt6`

Explanation:

`2x^2-10+7=0`
Add 3 on both sides to move the constants to the right side:
`2x^2-10+7+3=0+3`
`2x^2=3`
Divide both sides with 2:
`x^2=3/2`
`x=+-sqrt(3/2)=+-1/2sqrt3sqrt2=+-1/2sqrt6`
(if we multiply the right side with `sqrt2/sqrt2`)

Answer 2

`x = (5 pm sqrt(11))/2`

Explanation:

if `2x^2 -10x + 7 = 0`

You can use the quadratic formula:

`ax^2 +bx + c = 0`

`=> x = ( -bpmsqrt(b^2 - 4ac) ) / (2a)`

In this case `a = 2` , `b = -10` , `c = 7`

`=> x =( - (-10 ) pm sqrt( (-10)^2 - (4 * 2 * 7 ) ) ) / ( 2 * 2)`

`=> x = (10 pm sqrt(44)) / 4`

`=> x= (10 pm 2sqrt(11) )/ 4`

`color(red)(=> x = (5 pm sqrt(11))/2`