What are the solutions to the equation (r-3)(r+13)=-19 ?

The solutions to the equation (r-3)(r+13)=-19 are r=-5+3sqrt(5),r=-5-3sqrt(5)

Find the zeros of (r-3)(r+13)=-19

The zeros of (r-3)(r+13)=-19 are r=-5+3sqrt(5),r=-5-3sqrt(5)

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(r - 3) (r + 13) = - 19

r = - 5 + 35, r = - 5 - 35

Decimal Notation

r = 1.70820 \dots, r = - 11.70820 \dots

Solution steps

(r - 3) (r + 13) = - 19

Expand

(r - 3) (r + 13) : r^{2} + 10 r - 39

r^{2} + 10 r - 39 = - 19

Move

19

to the left side

r^{2} + 10 r - 20 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 20) = 65

r_{1, 2} = \frac{- 10 \pm 6 5}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 10 + 6 5}{2 \cdot 1}, r_{2} = \frac{- 10 - 6 5}{2 \cdot 1}

r = \frac{- 10 + 6 5}{2 \cdot 1} : - 5 + 35

r = \frac{- 10 - 6 5}{2 \cdot 1} : - 5 - 35

The solutions to the quadratic equation are:

r = - 5 + 35, r = - 5 - 35

- 5 (x + 1) (x - 9) = 0

1100 - q^{2} = 300 + q^{2}

\frac{( x + 3 ) ( x + 1 )}{2} = 12

y^{2} = - 7 y

y^{2} = - 28

What are the solutions to the equation (r-3)(r+13)=-19 ?
The solutions to the equation (r-3)(r+13)=-19 are r=-5+3sqrt(5),r=-5-3sqrt(5)
Find the zeros of (r-3)(r+13)=-19
The zeros of (r-3)(r+13)=-19 are r=-5+3sqrt(5),r=-5-3sqrt(5)