What are the solutions to the equation 12x^2+57x+45=0 ?

The solutions to the equation 12x^2+57x+45=0 are x=-1,x=-15/4

Find the zeros of 12x^2+57x+45=0

The zeros of 12x^2+57x+45=0 are x=-1,x=-15/4

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12 x^{2} + 57 x + 45 = 0

x = - 1, x = - \frac{15}{4}

Decimal Notation

x = - 1, x = - 3.75

Solution steps

12 x^{2} + 57 x + 45 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 57 \pm 5 7 ^{2} - 4 \cdot 12 \cdot 45}{2 \cdot 12}

5 7^{2} - 4 \cdot 12 \cdot 45 = 33

x_{1, 2} = \frac{- 57 \pm 33}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 57 + 33}{2 \cdot 12}, x_{2} = \frac{- 57 - 33}{2 \cdot 12}

x = \frac{- 57 + 33}{2 \cdot 12} : - 1

x = \frac{- 57 - 33}{2 \cdot 12} : - \frac{15}{4}

The solutions to the quadratic equation are:

x = - 1, x = - \frac{15}{4}

(x + 5)^{2} = 5

0.7 x^{2} + 0.21 x - 0.03 = 0

4 (x - 9) (5 x + 1) = 0

x, y^{3} + x y + x^{2} = 3

9 - x^{2} - 1 = - 2 x (x - 1)

What are the solutions to the equation 12x^2+57x+45=0 ?
The solutions to the equation 12x^2+57x+45=0 are x=-1,x=-15/4
Find the zeros of 12x^2+57x+45=0
The zeros of 12x^2+57x+45=0 are x=-1,x=-15/4