What are the solutions to the equation 4x^2+12x-27=13 ?

The solutions to the equation 4x^2+12x-27=13 are x=2,x=-5

Find the zeros of 4x^2+12x-27=13

The zeros of 4x^2+12x-27=13 are x=2,x=-5

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4 x^{2} + 12 x - 27 = 13

x = 2, x = - 5

Solution steps

4 x^{2} + 12 x - 27 = 13

Move

13

to the left side

4 x^{2} + 12 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 ( - 40 )}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 (- 40) = 28

x_{1, 2} = \frac{- 12 \pm 28}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 12 + 28}{2 \cdot 4}, x_{2} = \frac{- 12 - 28}{2 \cdot 4}

x = \frac{- 12 + 28}{2 \cdot 4} : 2

x = \frac{- 12 - 28}{2 \cdot 4} : - 5

The solutions to the quadratic equation are:

x = 2, x = - 5

z^{2} + 24 z + 140 = 0

98 x^{2} - 2 = 0

6 x^{2} + 2 = 34

\frac{( x + 4 ) ( x - 2 )}{4} = 10

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

What are the solutions to the equation 4x^2+12x-27=13 ?
The solutions to the equation 4x^2+12x-27=13 are x=2,x=-5
Find the zeros of 4x^2+12x-27=13
The zeros of 4x^2+12x-27=13 are x=2,x=-5