# math

posted by Jimmy

solve 2x^2+3x+8=0 and express the solutions in a+bi form.

Let's use the quadratic formula to solve for x and express those solutions in a+bi form.

x = [-b + or - sqrt(b^2 - 4ac)]/2a

Note: sqrt = square root.

a = 2, b = 3, and c = 8 from your problem.

Therefore:
x = [-3 + or - sqrt(3^2 - 4*2*8)]/2*2
x = [-3 + or - sqrt(-55)]4
x = [-3 + or - i sqrt(55)]/4

Solutions:
x = [-3 + i sqrt(55)]/4 = -3/4 + [sqrt(55)/4]i
x = [-3 - i sqrt(55)]/4 = -3/4 - [sqrt(55)/4]i

I hope this will help.

## Similar Questions

1. ### ALGEBRA 1

Solve by whatever Method 1. X^2 +8X = -16 X^2 +8X +16 = -16 +16 (X +4)^2 =0 X+4 =+/- SQRT 0 X=-4+/- 0 X=-4, X=4 2. 3x^2-2x-5=0 X=-b+/- sqrt (b)^2-4ac 2a a=3 ,b=-4,c=-5 X=-(-2)+/- Sqrt (-2)^2-4(3)(-5) 2(3) X=2+/sqrt64 = 2+/-8 6 X= 2+8=10/6=1 …
2. ### maths

which solution is the correct for this equation 4x^3 - 3x = 15 -1.8, -1.7, -1.6, -1.5, 1.5, 1.6, 1.7, 1.8 ?
3. ### Math

Use the quadratic formula to solve each of the following quadratic equations... 1. 2x^2-5x=3 2. 3x^2-2x+1=0 Rearrange the equation in quadratic formula form. 2x^2 -5x -3 = 0 Then use the formula. Tell me what you don't understand about …
4. ### Calculus - Second Order Differential Equations

Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 y'(0)=4, …
5. ### Calculus - Second Order Differential Equations

Posted by COFFEE on Monday, July 9, 2007 at 9:10pm. download mp3 free instrumental remix Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 …
6. ### Calculus

Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 …