How do you solve $10 x^{2} - 17 x = 63$ $10 x ^2 − 17 x = 63$ ?

Question

2051 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-04T13:52:50

$10 x^{2} - 17 x = 63$ $10x^2-17x=63$

$\to 10 x^{2} - 17 x - 63 = 0$ $rarr10x^2-17x-63=0$

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

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

In this case $a = 10, b = - 17, c = - 63$ $a=10,b=-17,c=-63$

$\to x = \frac{- (- 17) \pm \sqrt{{(- 17)}^{2} - 4 (10) (- 63)}}{2 (10)}$ $rarrx=(-(-17)+-sqrt((-17)^2-4(10)(-63)))/(2(10))$

$\to x = \frac{17 \pm \sqrt{289 - (- 2520)}}{20}$ $rarrx=(17+-sqrt(289-(-2520)))/(20)$

$\to x = \frac{17 \pm \sqrt{289 + 2520}}{20}$ $rarrx=(17+-sqrt(289+2520))/20$

$\to x = \frac{17 \pm \sqrt{2809}}{20}$ $rarrx=(17+-sqrt(2809))/20$

$\to x = \frac{17 \pm 53}{20}$ $rarrx=(17+-53)/20$

Now we can assume the two values of $x$ $x$ :

$\to x = \frac{17 + 53}{20}, \frac{17 - 53}{20}$ $rarrx=(17+53)/20,(17-53)/20$

$\to x = \frac{70}{20}, - \frac{36}{20}$ $rarrx=70/20,-36/20$

$\to x = 3 \frac{1}{2}, - 1 \frac{4}{5}$ $rarrx=3 1/2,-1 4/5$

How do you solve 10x2−17x=63 10 x ^2 − 17 x = 63 ?