What are the solutions to the equation 4.5x^2+32x=940 ?

The solutions to the equation 4.5x^2+32x=940 are x=(2(sqrt(4486)-16))/9 ,x=-(2(16+sqrt(4486)))/9

Find the zeros of 4.5x^2+32x=940

The zeros of 4.5x^2+32x=940 are x=(2(sqrt(4486)-16))/9 ,x=-(2(16+sqrt(4486)))/9

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4.5 x^{2} + 32 x = 940

x = \frac{2 ( 4486 - 16 )}{9}, x = - \frac{2 ( 16 + 4486 )}{9}

Decimal Notation

x = 11.32835 \dots, x = - 18.43946 \dots

Solution steps

4.5 x^{2} + 32 x = 940

Multiply both sides by

10

45 x^{2} + 320 x = 9400

Move

9400

to the left side

45 x^{2} + 320 x - 9400 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 320 \pm 32 0 ^{2} - 4 \cdot 45 ( - 9400 )}{2 \cdot 45}

32 0^{2} - 4 \cdot 45 (- 9400) = 204486

x_{1, 2} = \frac{- 320 \pm 20 4486}{2 \cdot 45}

Separate the solutions

x_{1} = \frac{- 320 + 20 4486}{2 \cdot 45}, x_{2} = \frac{- 320 - 20 4486}{2 \cdot 45}

x = \frac{- 320 + 20 4486}{2 \cdot 45} : \frac{2 ( 4486 - 16 )}{9}

x = \frac{- 320 - 20 4486}{2 \cdot 45} : - \frac{2 ( 16 + 4486 )}{9}

The solutions to the quadratic equation are:

x = \frac{2 ( 4486 - 16 )}{9}, x = - \frac{2 ( 16 + 4486 )}{9}

11600 = \frac{n}{2} (200 + (n - 1) 20)

(x + 1) (x - 2) = 10

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

(x + 1) (x + 2) + 2 = 0

100 \cdot (1 - ((1 - x) \cdot (1 - x))) = 2

What are the solutions to the equation 4.5x^2+32x=940 ?
The solutions to the equation 4.5x^2+32x=940 are x=(2(sqrt(4486)-16))/9 ,x=-(2(16+sqrt(4486)))/9
Find the zeros of 4.5x^2+32x=940
The zeros of 4.5x^2+32x=940 are x=(2(sqrt(4486)-16))/9 ,x=-(2(16+sqrt(4486)))/9