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This is a systems equation.

1. Bring all the variables to the same side (right) and constants to the same side (left).

This gives you : 2x^2 - 3x = 6
4x^2 + 5x = 6
It became +5x because the sign of the unit changes when it changes sides

2. Now you want to cancel out one of the variables on the right, either x^2 or x, because you want to subtract one equation from the other. Let's do x^2. In order to do this, equation (a) has to become 4x^2 - 6x = 12. You do this by multiplying equation (a) by 2.

2(2x^2 - 3x = 6)
4x^2 + 5x = 6

4x^2 - 6x = 12
4x^2 + 5x = 6

3. Now subtract the second equation from the first one,

4x^2 - 6x = 12
- 4x^2 + 5x = 6

The result is -6x - 5x = 6.
-11x = 6.

4. Now solve for x by dividing 6 by -11 to get
x= -6/11

Question a
2x² - 3x - 6 = 0
x = [ 3 ± √ (9 + 48 ) ] / 4
x = [ 3 ± √ 57 ] / 4

Question b
4x² + 5x - 6 = 0
x = [ - 5 ± √ (25 + 96 ) ] / 8
x = [ - 5 ± √ (121) ] / 8
x = [ - 5 ± 11 ] / 8
x = 6/8 , x = - 16/8
x = 3/4 , x = - 2

a)
2x^2 - 3x = 6
2x^2 - 3x - 6 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 2
b = -3
c = -6

x = [3 ±√(9 + 48)]/4
x = [3 ±√57]/4
x = [3 ±7.54]/4 (approx.)

x = [3 + 7.54]/4
x = 10.54/4
x = 2.635

x = [3 - 7.54]/4
x = -4.54/4
x = -1.135

∴ x = 2.635 , -1.135

= = = = = = = =

b)
4x^2 = 6 - 5x
4x^2 + 5x - 6 = 0
4x^2 + 8x - 3x - 6 = 0
4x(x + 2) - 3(x + 2) = 0
(x + 2)(4x - 3) = 0

x + 2 = 0
x = -2

4x - 3 = 0
4x = 3
x = 3/4 (0.75)

∴ x = -2 , 3/4 (0.75)

2x^2 - 3x - 6 = 0
x = [-b+/- sqrt(b^2-4ac) ]/2a
x = [3 +/- sqrt((-3)^2-4(2)(-6)) ]/2(2)
x = [3 +/- sqrt(9+48) ]/4
x = [3+/- sqrt(57) ]/4


4x^2 + 5x - 6 =0
x = [-b+/- sqrt(b^2-4ac) ]/2a
x = [-5+/- sqrt((5)^2-4(4)(-6)) ]/2(4)
x = [-5 +/- sqrt(25 + 96) /8
x = [-5+/- sqrt(121) ]/8
x = [-5+/-11 ]/8

x = -2 or 3/4

You have a couple of quadratics, and the best way to solve quadratics is to put in general form (with a zero on the right):

2x^2 - 3x = 6
2x^2 - 3x - 6 = 0

Since we don't have a neat factorisation, we need to use the quadratic formula to get this:

x = [3 +/- sqrt(57)]/4

For b:

4x^2 = 6 - 5x
4x^2 + 5x - 6 = 0

This one can be factorised (or you can use the formula):

(4x - 3)(x + 2) = 0
x = -2, 3/4

a.2x^2-3x=6
2x^2-3x-6=0 (set the equation equal to zero by subtracting 6 from both sides)
Then all you have to do is factor the equation or use the quadratic formula

use the same steps for the next question

these are just hidden quadratics

For instance,

a) 2x^2-3x=6
2x^2-3x-6=0
3 +/- sqrt(9+48)/4

x= 3 +/- sqrt(57)/4

b)4x^2+5x-6=0
(4x^2+8x)+(-3x-6)
4x(x+2)-3(x+2)
(4x-3)(x+2)
x=3/4 x= -2