2x^2+9x=-18?

/
2x^2+9x+18=0
x=-3,-3/2

Factoring:
2x² + 9x = - 18
x² + 9/2x = - 9
x² + 9/4x = - 9 + (9/4)²
x² + 9/4x = - 144/16 + 81/16
(x + 9/4) = - 63/16
x + 9/4 = - √63/4
x + ([9 + √63]/4) = 0
x = - (9 + √63)/4

Completing the square method:
2x^2 +9x = -18
2x^2 +9x +18 = 0 +18 at both sides
x^2 +9x/2 +9 =0 divide 2 at both sides
x^2 +9x/2 +(9/4)^2 -(9/4)^2 +9 = 0
(x +9/4)^2 +63/16 = 0 factor and evalute
(x +9/4)^2 = -63/16 -63/16 at both sides
From there, it is complex numbers, so we need to solve it by existing imaginary number. Thus,
x +9/4 = +/- square root of -63/16
x +9/4 = +/- (square root of 9*7*(-1))/4
x +9/4 = +/- 3(square root of 7)i/4
x = -9/4 +/- 3(square root of 7)i/4
where i is imaginary number by defination i = square root of (-1).
Finally, the complex roots are
-9/4 + 3(square root of 7)i/4 and -9/4 - 3(square root of 7)i/4

2x^2 + 9x = -18
2x^2 + 9x + 18 = 0
x = [-b ±√(b^2 - 4ac)]/2a (quadratic discriminant)

a = 2
b = 9
c = 18

x = [-9 ±√(81 - 144)]/4
x = [-9 ±√-63]/4 (imaginary number)
(no real roots)

2x² + 9x + 18 = 0
x = [ - 9 ± √ (81 - 144 ) ] / 4
x = [ - 9 ± √ (- 63 ) ] / 4
x = [ - 9 ± √ (63 i² ) ] / 4
x = [ - 9 ± i √ (63) ] / 4

2x^2+9x+18=0
x=(-9+/-Sqrt(9^2-2*2*18))/(2*2)
= (-9+/-3)/4
= -3 and -3/2

D=-63
X= -2.25+1.98i, -2.25-1.98i
(complex conjugates)

Here:

2x^2 + 9x = -18
Divide each side by 2.

x^2 + 4.5x = -9
Add 9 to each side.

x^2 + 4.5x + 9 = 0

Now, subsitute the values for ax^2 + bx + c = 0
where a = 1, b = 4.5, and c = 9

for

x= (-b +- sqrt[b^2 - 4ac])/2a
so,
x= (-4.5 +- sqrt[4.5^2 - 4{1}{9}])/2{1}

and, you get

-4.5 +- sqrt(-15.75)/2

which will give us complex roots (an imaginary answer)

so, x = -2.25 + 1.983i
or, x = -2.25 - 1.983i

Sincerely,
R.T.K.