What are the solutions to the equation r^2-3r-40=0 ?

The solutions to the equation r^2-3r-40=0 are r=8,r=-5

Find the zeros of r^2-3r-40=0

The zeros of r^2-3r-40=0 are r=8,r=-5

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r^{2} - 3 r - 40 = 0

r = 8, r = - 5

Solution steps

r^{2} - 3 r - 40 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 40 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 40) = 13

r_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 3 ) + 13}{2 \cdot 1}, r_{2} = \frac{- ( - 3 ) - 13}{2 \cdot 1}

r = \frac{- ( - 3 ) + 13}{2 \cdot 1} : 8

r = \frac{- ( - 3 ) - 13}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

r = 8, r = - 5

x - 9 \leq 2 (9 - x)

1 - 6 x + 9 x^{2} - 4 y^{2}

\frac{m - 2}{3} < - 2 or 4 m + 3 > 15

45 x^{2} - 8 x y - 4 y^{2}

2 r^{2} + r - 4 = 0

What are the solutions to the equation r^2-3r-40=0 ?
The solutions to the equation r^2-3r-40=0 are r=8,r=-5
Find the zeros of r^2-3r-40=0
The zeros of r^2-3r-40=0 are r=8,r=-5