What are the solutions to the equation 12x+3=-4x^2+20 ?

The solutions to the equation 12x+3=-4x^2+20 are x=-(3+sqrt(26))/2 ,x=(sqrt(26)-3)/2

Find the zeros of 12x+3=-4x^2+20

The zeros of 12x+3=-4x^2+20 are x=-(3+sqrt(26))/2 ,x=(sqrt(26)-3)/2

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12 x + 3 = - 4 x^{2} + 20

x = - \frac{3 + 26}{2}, x = \frac{26 - 3}{2}

Decimal Notation

x = - 4.04950 \dots, x = 1.04950 \dots

Solution steps

12 x + 3 = - 4 x^{2} + 20

Switch sides

- 4 x^{2} + 20 = 12 x + 3

Move

3

to the left side

- 4 x^{2} + 17 = 12 x

Move

12 x

to the left side

- 4 x^{2} + 17 - 12 x = 0

Write in the standard form

a x^{2} + b x + c = 0

- 4 x^{2} - 12 x + 17 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 4 ) \cdot 17}{2 ( - 4 )}

(- 12)^{2} - 4 (- 4) \cdot 17 = 426

x_{1, 2} = \frac{- ( - 12 ) \pm 4 26}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4 26}{2 ( - 4 )}, x_{2} = \frac{- ( - 12 ) - 4 26}{2 ( - 4 )}

x = \frac{- ( - 12 ) + 4 26}{2 ( - 4 )} : - \frac{3 + 26}{2}

x = \frac{- ( - 12 ) - 4 26}{2 ( - 4 )} : \frac{26 - 3}{2}

The solutions to the quadratic equation are:

x = - \frac{3 + 26}{2}, x = \frac{26 - 3}{2}

(x + 3) 2 = 8 (x - 2) + x^{2}

x^{2} + 13 x + 24 = 0

2 b^{2} + 40 = 0

(\frac{5}{13})^{2} + x^{2} = 1

6 s^{2} + 18 s = 0

What are the solutions to the equation 12x+3=-4x^2+20 ?
The solutions to the equation 12x+3=-4x^2+20 are x=-(3+sqrt(26))/2 ,x=(sqrt(26)-3)/2
Find the zeros of 12x+3=-4x^2+20
The zeros of 12x+3=-4x^2+20 are x=-(3+sqrt(26))/2 ,x=(sqrt(26)-3)/2