2x^2+9x-5=0 solve by completing the square?

2(x^2+ 9/2 x - 5/2) = 0

2((x + 9/4)^2 - 7.5625) = 0

(x + 9/4)^2 = 7.5625

x + 9/4 = +/- sqrt(7.5625)

x = -9/4 +/- 11/4

x = -5 or 1/2

you do not could characteristic a relentless in the past winding up the sq. as you have achieved there and since the above answerer have achieved, basically green human beings do this. complete the sq. for this expression: x² + 4x = 21 (x + 2)² - 4 = 21 (x + 2)² = 25 x + 2 = ±5 x = -7 OR x = 3 x = {-7, 3}

2x^2+9x-5=0

Step 1:

Move 5 to other side: 2x^2+9x=5

Step 2:

get rid of a usually helps. x^2+4.5=2.5

Step 3:

use (b/2)^2 to get (4.5/2)^2= 5.0625

Step 4:

plug it in. x^2+4.5x+5.0625=7.5625

or in fractions: x^2+4 1/2+ 5 1/16= 7 9/16

Step 5:

factor the first side for

(x+2 1/4)^2 or (x+2.25)^2

Step 6:

square root both sides

(x+2 1/4)= 2 3/4

Step 7:

Now do x= -2.25+2.75 and x=-2.25-2.75

x= 0.5, -5

ps. sepia and Big D did it by factoring not completeing the square.

2x^2 + 9x - 5 = 0

(x + 5)(2x - 1) = 0

x = 1/2 or -5

(2x-1) (x+5)=0

2x-1=0

2x=1

x=1/2

x+5=0

x=-5

x=1/2, -5