What are the solutions to the equation 2(5x+4)^2-40=-20 ?

The solutions to the equation 2(5x+4)^2-40=-20 are x=(sqrt(10)-4)/5 ,x=(-sqrt(10)-4)/5

Find the zeros of 2(5x+4)^2-40=-20

The zeros of 2(5x+4)^2-40=-20 are x=(sqrt(10)-4)/5 ,x=(-sqrt(10)-4)/5

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2 (5 x + 4)^{2} - 40 = - 20

x = \frac{10 - 4}{5}, x = \frac{- 10 - 4}{5}

Decimal Notation

x = - 0.16754 \dots, x = - 1.43245 \dots

Solution steps

2 (5 x + 4)^{2} - 40 = - 20

Move

40

to the right side

2 (5 x + 4)^{2} = 20

Divide both sides by

2

(5 x + 4)^{2} = 10

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

5 x + 4 = 10 : x = \frac{10 - 4}{5}

Solve

5 x + 4 = - 10 : x = \frac{- 10 - 4}{5}

The solutions to the quadratic equation are:

x = \frac{10 - 4}{5}, x = \frac{- 10 - 4}{5}

36 m^{2} - 24 m + 16 n^{2}

2 x^{2} - 10 = 0

p (9 p - 16) < 0

\frac{5}{3} \cdot 9

∣ n ∣ + 1 = 2

What are the solutions to the equation 2(5x+4)^2-40=-20 ?
The solutions to the equation 2(5x+4)^2-40=-20 are x=(sqrt(10)-4)/5 ,x=(-sqrt(10)-4)/5
Find the zeros of 2(5x+4)^2-40=-20
The zeros of 2(5x+4)^2-40=-20 are x=(sqrt(10)-4)/5 ,x=(-sqrt(10)-4)/5