If it's not what You are looking for type in the equation solver your own equation and let us solve it.
v^2+25v+15=0
a = 1; b = 25; c = +15;
Δ = b2-4ac
Δ = 252-4·1·15
Δ = 565
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{565}}{2*1}=\frac{-25-\sqrt{565}}{2} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{565}}{2*1}=\frac{-25+\sqrt{565}}{2} $
| -12=-8v+4(v=8) | | 6(x+2)=12(x-1) | | 8-6y=4y-3 | | 7g+14-5g=-1 | | y/10+4=-7 | | -19c=-17-20c | | 3r-2/5=5+2r/8 | | -7x+4-6=-8x-2 | | 2^(x+1)-2^(x-1)=24 | | 36=9d | | 7x-6=2+7x-3-5 | | 7x-18=6x+30 | | x=90+.5x | | X+11-6x=3x-5 | | 4^2x-3=1 | | 7(3y+4)=5(4y+5)+3 | | 5x-10+5x=50 | | 7r-29=34 | | |2x-3|-1=0 | | 4k=-3 | | X/y=19 | | 10x+6-7x=40 | | -4q^2+30q+1=0 | | 16=3w-7w | | -6(2x-4)=-4(3x-6) | | X+9/12=3/4x | | 3x-9/2=5(2x-8)-2(4x-1) | | 5(x+2)-5=3x+5+2x | | X-6x-3x=-11-5 | | 11c-2(c+1)=3(c-6) | | z/2+2=9-z/2 | | b–2=1 |