What are the solutions to the equation 40=t^2+3t ?

The solutions to the equation 40=t^2+3t are t=5,t=-8

Find the zeros of 40=t^2+3t

The zeros of 40=t^2+3t are t=5,t=-8

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40 = t^{2} + 3 t

t = 5, t = - 8

Solution steps

40 = t^{2} + 3 t

Switch sides

t^{2} + 3 t = 40

Move

40

to the left side

t^{2} + 3 t - 40 = 0

Solve with the quadratic formula

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

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

t_{1, 2} = \frac{- 3 \pm 13}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 3 + 13}{2 \cdot 1}, t_{2} = \frac{- 3 - 13}{2 \cdot 1}

t = \frac{- 3 + 13}{2 \cdot 1} : 5

t = \frac{- 3 - 13}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

t = 5, t = - 8

x^{4} - 4 x^{3} - x^{2} + 4 x

3 (x^{2} - 1) > - 8 x

(x + 1)^{2} = 7

3 x^{2} + 17 x + 20 = 0

∣ x ∣ \leq 20

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