What are the solutions to the equation (7-3x)(x+2)=4 ?

The solutions to the equation (7-3x)(x+2)=4 are x=-5/3 ,x=2

Find the zeros of (7-3x)(x+2)=4

The zeros of (7-3x)(x+2)=4 are x=-5/3 ,x=2

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(7 - 3 x) (x + 2) = 4

x = - \frac{5}{3}, x = 2

Decimal Notation

x = - 1.66666 \dots, x = 2

Solution steps

(7 - 3 x) (x + 2) = 4

Expand

(7 - 3 x) (x + 2) : - 3 x^{2} + x + 14

- 3 x^{2} + x + 14 = 4

Move

4

to the left side

- 3 x^{2} + x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 3 ) \cdot 10}{2 ( - 3 )}

1^{2} - 4 (- 3) \cdot 10 = 11

x_{1, 2} = \frac{- 1 \pm 11}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 1 + 11}{2 ( - 3 )}, x_{2} = \frac{- 1 - 11}{2 ( - 3 )}

x = \frac{- 1 + 11}{2 ( - 3 )} : - \frac{5}{3}

x = \frac{- 1 - 11}{2 ( - 3 )} : 2

The solutions to the quadratic equation are:

x = - \frac{5}{3}, x = 2

x^{2} - 6 x - 7 = 0

15000 = - 0.5 x^{2} + 40 x - 300

\frac{x ^{2}}{3} = \frac{11}{6} x + 1

(k + 2) x^{2} + 5 x - 3 = 0

(x + 2)^{2} - 36 = 0

What are the solutions to the equation (7-3x)(x+2)=4 ?
The solutions to the equation (7-3x)(x+2)=4 are x=-5/3 ,x=2
Find the zeros of (7-3x)(x+2)=4
The zeros of (7-3x)(x+2)=4 are x=-5/3 ,x=2