What are the solutions of the quadratic equation 4x^2+34 x+60=0?

Answer:Step-by-step explanation:Start by factoring out a 2 just to make the coefficients a bit smaller:[tex]0=2(2x^2+17x+30)[/tex]Then factor the remaining simplified quadratic. This is in standard form, and a= 2, b = 17 and c = 30.  We will find the product of a and c:  2 * 30 = 60.  Now when we find the factors of 60, some combination of 2 factors of 60 will have to add to be the linear term, 17.  The factors of 60 are:  1, 60; 2, 30; 3, 20; 4, 15; 5, 12; 6, 10.  Two of those will work:  20 - 3 = 17, and 12 + 5 = 17.  Let's start with the first set.  They figure into the factoring formula as follows:[tex]2x^2+20x-3x+30=0[/tex]Now group them together into groups of 2 without moving any of the terms:[tex](2x^2+20x)-(3x+30)[/tex]From each term factor out whatever is common between the coefficients and the variables:[tex]2x(x+10)-3(x-10)[/tex]If this pair of factors "works" what's inside each set of parenthesis should be EXACTLY the same.  They are not, so this particular set of factors didn't work.  Let's try the other set: 5, 12:[tex]2x^2+12x+5x+30=0[/tex]Group them together in groups of 2:[tex](2x^2+12x)+(5x+30)=0[/tex]and factor out what's common in each set of parenthesis:2x(x + 6) + 5(x + 6) = 0The parenthesis have exactly the same term inside them so this one "worked".  We will then factor out that set of parenthesis and put what's "left" inside another set of parenthesis and we will be done.(x + 6)(2x + 5) = 0The solutions are:x + 6 = 0 so x = -62x + 5 = 0 so 2x = -5 and [tex]x=-\frac{5}{2}[/tex]So there you go!