What are the solutions to the equation 4=a^2+3a ?

The solutions to the equation 4=a^2+3a are a=1,a=-4

Find the zeros of 4=a^2+3a

The zeros of 4=a^2+3a are a=1,a=-4

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4 = a^{2} + 3 a

a = 1, a = - 4

Solution steps

4 = a^{2} + 3 a

Switch sides

a^{2} + 3 a = 4

Move

4

to the left side

a^{2} + 3 a - 4 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

a_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- 3 + 5}{2 \cdot 1}, a_{2} = \frac{- 3 - 5}{2 \cdot 1}

a = \frac{- 3 + 5}{2 \cdot 1} : 1

a = \frac{- 3 - 5}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

a = 1, a = - 4

x^{2} = 2 5^{2} + 7^{2}

144 = x^{2} - 52

(50 - 2 x) (80 - 2 x) = 2000

- 2 x^{2} + 3 x = - 1

24 x^{2} - 50 x - 20 = 2 x

What are the solutions to the equation 4=a^2+3a ?
The solutions to the equation 4=a^2+3a are a=1,a=-4
Find the zeros of 4=a^2+3a
The zeros of 4=a^2+3a are a=1,a=-4