Respuesta :

calculista

Answer:

The two values of x are -2.5 and -3

Step-by-step explanation:

we have

[tex]2x^{2}+11x+15=0[/tex]

The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to

[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]

in this problem we have

[tex]a=2\\b=11\\c=15[/tex]

substitute in the formula

[tex]x=\frac{-11(+/-)\sqrt{11^{2}-4(2)(15)}} {2(2)}[/tex]

[tex]x=\frac{-11(+/-)\sqrt{1}} {4}[/tex]

[tex]x=\frac{-11(+/-)1} {4}[/tex]

[tex]x_1=\frac{-11(+)1} {4}=-2.5[/tex]

[tex]x_2=\frac{-11(-)1} {4}=-3[/tex]

therefore

The two values of x are -2.5 and -3

profantoniofonte

Answer:

[tex]S=\left \{ \frac{-5}{2}, -3\right \}[/tex]

Step-by-step explanation:

Solving it by factoring. Firstly by multiplying the parameter a (2) by c(15) =30 then find two numbers whose product is 30 and their sum is 11. Here we have: 6 and 5. 6x5=30 6+5=11. Rewrite the b parameter as 6x+5x replacing 11x as it follows, then factor by grouping:

[tex]2x^{2}+11x+15\\2x^{2}+6x+5x+15\Rightarrow (2x^{2}+6x)+(5x+15)\Rightarrow 2x(x+3)+5(x+3)\Rightarrow (2x+5)(x+3)[/tex]

Solving each factor separately as a linear equation to find x' and x'':

[tex]2x+5=0\Rightarrow 2x=-5\Rightarrow x=\frac{-5}{2}\\(x+3)=0\Rightarrow x=-3\\S=\left \{ \frac{-5}{2}, -3\right \}[/tex]

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