What are the solutions to the equation 4x^2=12-bx ?

The solutions to the equation 4x^2=12-bx are x=(-b+sqrt(b^2+192))/8 ,x=(-b-sqrt(b^2+192))/8

Find the zeros of 4x^2=12-bx

The zeros of 4x^2=12-bx are x=(-b+sqrt(b^2+192))/8 ,x=(-b-sqrt(b^2+192))/8

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4 x^{2} = 12 - b x

x = \frac{- b + b ^{2} + 192}{8}, x = \frac{- b - b ^{2} + 192}{8}

Solution steps

4 x^{2} = 12 - b x

Move

b x

to the left side

4 x^{2} + b x = 12

Move

12

to the left side

4 x^{2} + b x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- b \pm b ^{2} - 4 \cdot 4 ( - 12 )}{2 \cdot 4}

Simplify

b^{2} - 4 \cdot 4 (- 12) : b^{2} + 192

x_{1, 2} = \frac{- b \pm b ^{2} + 192}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- b + b ^{2} + 192}{2 \cdot 4}, x_{2} = \frac{- b - b ^{2} + 192}{2 \cdot 4}

x = \frac{- b + b ^{2} + 192}{2 \cdot 4} : \frac{- b + b ^{2} + 192}{8}

x = \frac{- b - b ^{2} + 192}{2 \cdot 4} : \frac{- b - b ^{2} + 192}{8}

The solutions to the quadratic equation are:

x = \frac{- b + b ^{2} + 192}{8}, x = \frac{- b - b ^{2} + 192}{8}

- 4 x^{2} + 7 x - 2 = 0

2 (3 x + 1)^{2} - 32 = 0

2 w^{2} - 7 w - 4 = 0

(x + 3)^{2} + 4 (x + 2)^{2} = 2 5^{2}

5 = 3 x^{2}

What are the solutions to the equation 4x^2=12-bx ?
The solutions to the equation 4x^2=12-bx are x=(-b+sqrt(b^2+192))/8 ,x=(-b-sqrt(b^2+192))/8
Find the zeros of 4x^2=12-bx
The zeros of 4x^2=12-bx are x=(-b+sqrt(b^2+192))/8 ,x=(-b-sqrt(b^2+192))/8