# Solve the following: a^2-6a+2=0 and 2x^2+3x=2

Asked on by peuco7

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equations to be solved are a^2 - 6a + 2 = 0 and 2x^2 + 3x = 2

a^2 - 6a + 2 = 0

The roots of the equation ax^2 + bx + c = 0 are `x1 = (-b + sqrt(b^2 - 4ac))/(2a)` and `x2 = (-b - sqrt(b^2 - 4ac))/(2a)`

=> a1 = `3 + sqrt 7` and a2 = `3 - sqrt 7`

2x^2 + 3x = 2

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

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

=> 2x(x + 2) - 1(x + 2) = 0

=> (2x - 1)(x + 2) = 0

=> x = `1/2` and x = -2

The solutions of a^2 - 6x + 2 = 0 are `3 + sqrt 7` and `3 - sqrt 7` and the solutions of 2x^2 + 3x = 2 are` 1/2` and -2.

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

Posted on

a^2-6a+2=0 and 2x^2+3x=2

a^2-6a+2=0

b= -6a divide by 2 which is -3 Square it (9)

add it to both sides

-6a + 9 + a2 = -2 + 9

combine like terms

-2 + 9 = 7

9 + -6a + a2 = 7

Factor a and c

`sqrt((a + -3)^2)=+-sqrt(7) `

`(a-3)=+-sqrt(7)`

`a=3-sqrt(7) `

`a=3+sqrt(7) `

a= 2.645751311 and -2.645751311.

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