What are the solutions to the equation (5n+1)^2+2(5n+1)-3=0 ?

The solutions to the equation (5n+1)^2+2(5n+1)-3=0 are n=0,n=-4/5

Find the zeros of (5n+1)^2+2(5n+1)-3=0

The zeros of (5n+1)^2+2(5n+1)-3=0 are n=0,n=-4/5

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(5 n + 1)^{2} + 2 (5 n + 1) - 3 = 0

n = 0, n = - \frac{4}{5}

Decimal Notation

n = 0, n = - 0.8

Solution steps

(5 n + 1)^{2} + 2 (5 n + 1) - 3 = 0

Expand

(5 n + 1)^{2} + 2 (5 n + 1) - 3 : 25 n^{2} + 20 n

25 n^{2} + 20 n = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 25 \cdot 0}{2 \cdot 25}

2 0^{2} - 4 \cdot 25 \cdot 0 = 20

n_{1, 2} = \frac{- 20 \pm 20}{2 \cdot 25}

Separate the solutions

n_{1} = \frac{- 20 + 20}{2 \cdot 25}, n_{2} = \frac{- 20 - 20}{2 \cdot 25}

n = \frac{- 20 + 20}{2 \cdot 25} : 0

n = \frac{- 20 - 20}{2 \cdot 25} : - \frac{4}{5}

The solutions to the quadratic equation are:

n = 0, n = - \frac{4}{5}

81 m^{2} + 48 m

∣ x + 8 ∣ = 4

9^{2} + b^{2} = 1 6^{2}

∣ x ∣ \geq 6

x + 8 < - 7

What are the solutions to the equation (5n+1)^2+2(5n+1)-3=0 ?
The solutions to the equation (5n+1)^2+2(5n+1)-3=0 are n=0,n=-4/5
Find the zeros of (5n+1)^2+2(5n+1)-3=0
The zeros of (5n+1)^2+2(5n+1)-3=0 are n=0,n=-4/5