can you solve this equation?

Since both are equations are set equal to y, then:

12x^2 - 5 = 7 - 7x

Bring everything to the left side:
12x^2 - 5 - 7 + 7x = 0

Simplify:
12x^2 + 7x - 12 = 0


Factor:
(4x - 3)(3x + 4) = 0

Set each factor equal to 0 and solve for x
4x - 3 = 0
4x = 3
x = 3/4

and

3x + 4 = 0
3x = -4
x = -4/3

Solution: x = -4/3 or 3/4

Hope this helps you
:o)
d

12 x ² - 5 = 7 - 7 x
12 x ² + 7x - 12 = 0
(4 x - 3)(3x + 4) = 0
x = 3/4 , x = - 4/3
y = 7 - 21/4 , y = 7 + 28/3
y = 7/4 , y = 49/3

(3/4,7/4) , (- 4/3,49/3)

y = 12x^2 - 5 (solve by using substitution)
y = 7 - 7x

y = 12x^2 - 5
7 - 7x = 12x^2 - 5
12x^2 + 7x - 5 - 7 = 0
12x^2 + 7x - 12 = 0
12x^2 + 16x - 9x - 12 = 0
(12x^2 + 16x) - (9x + 12) = 0
4x(3x + 4) - 3(3x + 4) = 0
(3x + 4)(4x - 3) = 0

3x + 4 = 0
3x = -4
x = -4/3

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

y = 7 - 7x
y = 7 - 7(-4/3)
y = 7 - (-28/3)
y = 7 + 28/3
y = 21/3 + 28/3
y = 49/3

y = 7 - 7(3/4)
y = 7 - 21/4
y = 28/4 - 21/4
y = 7/4 (1.75)

∴ [x = -4/3 , y = 49/3] , [x = 3/4 (0.75) , y = 7/4 (1.75)]

12x^2-5=7-7x,12x^2+7x-12=0,x=-7+-(49+576)^1/2/24
x=0.75 & -4/3
y=1.75 & -2.3333

y=12x^2-5
y=7-7x
now LHS=RHS
so equate both the equations
12x^2-5=7-7x
12x^2-5-7+7x=0
12x^2+7x-12=0
solve for x
x= -4/3 or x=3/4

7-7x = 12x^2-5

12x^2 + 7x - 12 = 0

(4x - 3)(3x + 4) = 0

x = 3/4 or x = - 4/3

y = 7 - 21/4 , y = 7 + 28/3

y = 7/4 or y = 49 / 3