How do you solve $t^{2} + 5 t - 24 = 0$ $t^2 + 5t - 24 = 0$ ?

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Answer 1 · 2016-03-26T06:43:15

$x = 3, - 8$ $x=3,-8$

$t^{2} + 5 t - 24 = 0$ $color(blue)(t^2+5t-24=0$

This is a Quadratic equation (in form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ )

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)$

Where

$a = 1, b = 5, c = - 24$ $color(red)(a=1,b=5,c=-24$

$\to x = \frac{- 5 \pm \sqrt{5^{2} - 4 (1) (- 24)}}{2 (1)}$ $rarrx=(-5+-sqrt(5^2-4(1)(-24)))/(2(1))$

$\to x = \frac{- 5 \pm \sqrt{25 - (- 96)}}{2}$ $rarrx=(-5+-sqrt(25-(-96)))/(2)$

$\to x = \frac{- 5 \pm \sqrt{25 + 96}}{2}$ $rarrx=(-5+-sqrt(25+96))/(2)$

$\to x = \frac{- 5 \pm \sqrt{121}}{2}$ $rarrx=(-5+-sqrt(121))/(2)$

$\to x = \frac{- 5 \pm 11}{2}$ $rarrx=(-5+-11)/(2)$

Now we have $2$ $2$ solutions

$x = \frac{- 5 + 11}{2} = \frac{6}{2} = 3$ $color(purple)(x=(-5+11)/(2)=6/2=3$

$x = \frac{- 5 - 11}{2} = - \frac{16}{2} = - 8$ $color(indigo)(x=(-5-11)/(2)=-16/2=-8$

$:. color(blue)( ul bar |x=3,-8|$

How do you solve t^2 + 5t - 24 = 0 t2+5t−24=0t^2 + 5t - 24 = 0 ?