Quadratic Equations (Easy)

Using the quadratic formula

$x=\dfrac{20}{2}\pm\dfrac{\sqrt{(-20)^2-4(1)(-69)}}{2}=10\pm\dfrac{\sqrt{400+276}}{2}=10\pm13$

$\implies$ $x=10+13=23$

$\implies$ $x=10-13=-3$

Substituting the values of $x$ into the equation, we have

when $x=23$

$23^2-20(23)-69=0$

$529-460-69=0$

$69-69=0$

$0=0$ (The statement is true)

when $x=-3$

$(-3)^2-20(-3)-69=0$

$9+60-69=0$

$69-69=0$

$0=0$ (The statement is true)

Therefore, the roots are $23$ and $-3$ .

The normal equation looked like $x^2-20x-69=0$ . But we will change the subtraction to the addition of the opposite to make it look like $x^2 +(-20x)+(-69)=0$ .he first number of the equation is inserted into A (in this case 1). The second number inserted for b (-20). And the third number for C (-69). In the end we put all the numbers in the quadratic formula to make: $\frac{(-20)\pm\sqrt{(-20)^2}-4*1*(-69)}{2*1} = 23,-3$

$x ^ { 2 } - 20x - 69 = 0 \rightarrow ( x + 3 ) ( x - 23 ) = 0 \Rightarrow x = -3, 23$ .