Answer:
1. The equation has 2 solutions 3. p = 0, p = -2
Step-by-step explanation:
1. The quadratic equation is ax^2+bx+c=0 [tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
in the equation a = 2, b = 3 and c = 5
so the equation would be [tex]x_{1,\:2}=\frac{-(-3)+ \sqrt{(-3)^2-4*2*5}}{2*2}[/tex]
[tex]x_{1,\:2}=\frac{3+\sqrt{9-4*2*5}}{2*2}[/tex]
simplify
[tex]3+\sqrt{31}i[/tex]
divide by 4
[tex]\frac{3}{4} +\frac{\sqrt{31} }{4}i[/tex] or [tex]x=\frac{-\left(-3\right)-\sqrt{\left(-3\right)^2-4\cdot \:2\cdot \:5}}{2\cdot \:2}:\quad \frac{3}{4}-i\frac{\sqrt{31}}{4}[/tex]
3. 4p(5p + 10) = 0 Using the Zero Factor Principle aka. zero-product property If a*b=0 then a=0 or b=0 or both a=0 and b=0
p = 0
5p + 10 = 0
subtract ten from both sides
5p = -10
divide by 5
p = -2
