What are the solutions to the equation 3x(x-5)+(x-3)(x-2)=-18 ?

The solutions to the equation 3x(x-5)+(x-3)(x-2)=-18 are x=3,x=2

Find the zeros of 3x(x-5)+(x-3)(x-2)=-18

The zeros of 3x(x-5)+(x-3)(x-2)=-18 are x=3,x=2

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3 x (x - 5) + (x - 3) (x - 2) = - 18

x = 3, x = 2

Solution steps

3 x (x - 5) + (x - 3) (x - 2) = - 18

Expand

3 x (x - 5) + (x - 3) (x - 2) : 4 x^{2} - 20 x + 6

4 x^{2} - 20 x + 6 = - 18

Move

18

to the left side

4 x^{2} - 20 x + 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 4 \cdot 24}{2 \cdot 4}

(- 20)^{2} - 4 \cdot 4 \cdot 24 = 4

x_{1, 2} = \frac{- ( - 20 ) \pm 4}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 4}{2 \cdot 4}, x_{2} = \frac{- ( - 20 ) - 4}{2 \cdot 4}

x = \frac{- ( - 20 ) + 4}{2 \cdot 4} : 3

x = \frac{- ( - 20 ) - 4}{2 \cdot 4} : 2

The solutions to the quadratic equation are:

x = 3, x = 2

(x + 3)^{2} = 6 x + 34

8 r^{2} + 4 r - 1 = 0

x, x^{2} y - 3 y + 2 = 0

\frac{3}{2} x^{2} + 11 = 59

(5 x + 24) (4 x + 37) = 0

What are the solutions to the equation 3x(x-5)+(x-3)(x-2)=-18 ?
The solutions to the equation 3x(x-5)+(x-3)(x-2)=-18 are x=3,x=2
Find the zeros of 3x(x-5)+(x-3)(x-2)=-18
The zeros of 3x(x-5)+(x-3)(x-2)=-18 are x=3,x=2