Complete the square -3x + 2x^2 = -7?

First divide by 2 giving x^2 - (3/2)x= -7/2 then add (3/4)^2 giving

x^2-(3/2)x+9/16=-7/2+9/16=-47/16 and taking roots gives

x-3/4=+/- sqrt(-47)/4 which gives your answer. Well done!

- 3x + 2x² = - 7

2x² - 3x + 7 = 0

2[x² - (3/2)x] + 7 = 0

2[x² - (3/2)x + (3/4)² - (3/4)²] + 7 = 0

2[x² - (3/2)x + (3/4)²] - 2(3/4)² + 7 = 0

2[x - (3/4)]² - 2(9/16) + 7 = 0

2[x - (3/4)]² - (9/8) + (56/8) = 0

2[x - (3/4)]² - (47/8) = 0

2[x - (3/4)]² - 2(47/16) = 0

[x - (3/4)]² - (47/16) = 0

[x - (3/4)]² = 47/16

x - (3/4) = ± (√47)/4

First case: x - (3/4) = + (√47)/4 → x = (3/4) + (√47)/4 → x = (3 + √47)/4

Second case: x - (3/4) = - (√47)/4 → x = (3/4) - (√47)/4 → x = (3 - √47)/4

→ Solution = { (3 - √47)/4 ; (3 + √47)/4 }

-3x + 2x^2 = -7..............rearrange

2x^2 - 3x = -7................divide by the coefficient of the squared term

x^2 - 3/2 x = -7/2 ...........add half the square of the coefficient of the nonsquared term to both sides

x^2 - 3/2 x + 9/16 = -28/4 + 9/16...Combine like terms & rewrite polynomial\

(x - 3/4)^2 = -17/16

to solve for x, take square root and add 3/4 you get (3/4) +/- (sqrt(17)/4)i

2x^2-3x=-7

or 2x^2-3x+7=0

orx=(3+sqrt(47i^2))/4 or x=(3-sqrt(47i^2))/4

thus x=(3/4)+/-(sqrt47i)/4

as i^2=-1