What are the solutions to the equation 4x+6x(x-5)+12=x^2+7 ?

The solutions to the equation 4x+6x(x-5)+12=x^2+7 are x=5,x= 1/5

Find the zeros of 4x+6x(x-5)+12=x^2+7

The zeros of 4x+6x(x-5)+12=x^2+7 are x=5,x= 1/5

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4 x + 6 x (x - 5) + 12 = x^{2} + 7

x = 5, x = \frac{1}{5}

Decimal

x = 5, x = 0.2

Solution steps

4 x + 6 x (x - 5) + 12 = x^{2} + 7

Expand

4 x + 6 x (x - 5) + 12 : 6 x^{2} - 26 x + 12

6 x^{2} - 26 x + 12 = x^{2} + 7

Move

7

to the left side

6 x^{2} - 26 x + 5 = x^{2}

Move

x^{2}

to the left side

5 x^{2} - 26 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 26 ) \pm ( - 26 ) ^{2} - 4 \cdot 5 \cdot 5}{2 \cdot 5}

(- 26)^{2} - 4 \cdot 5 \cdot 5 = 24

x_{1, 2} = \frac{- ( - 26 ) \pm 24}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 26 ) + 24}{2 \cdot 5}, x_{2} = \frac{- ( - 26 ) - 24}{2 \cdot 5}

x = \frac{- ( - 26 ) + 24}{2 \cdot 5} : 5

x = \frac{- ( - 26 ) - 24}{2 \cdot 5} : \frac{1}{5}

The solutions to the quadratic equation are:

x = 5, x = \frac{1}{5}

x^{2} + 16 x - 6 = 0

x^{2} + 6 x + 58 = 0

t, d = v t + \frac{a t ^{2}}{2}

6 x^{2} + 14 = - 25 x

- 16 x^{2} + 6 x + 22 = 5

What are the solutions to the equation 4x+6x(x-5)+12=x^2+7 ?
The solutions to the equation 4x+6x(x-5)+12=x^2+7 are x=5,x= 1/5
Find the zeros of 4x+6x(x-5)+12=x^2+7
The zeros of 4x+6x(x-5)+12=x^2+7 are x=5,x= 1/5