If it's not what You are looking for type in the equation solver your own equation and let us solve it.
42y^2+28y-14=0
a = 42; b = 28; c = -14;
Δ = b2-4ac
Δ = 282-4·42·(-14)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3136}=56$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-56}{2*42}=\frac{-84}{84} =-1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+56}{2*42}=\frac{28}{84} =1/3 $
| 17x=680 | | –10x=150 | | 3x-(+14)=180 | | 48+8t-t^=0 | | 120–600x=150 | | P(x)=95,000+200,000 | | 42-x=6 | | t^-8t-48=0 | | 9m^2-66m+21=0 | | 8y+5=4y+13 | | 12=48+8t-t^ | | 10x-23=25-14x | | 179=c+149 | | 12=48+8-t^ | | 12x+3=5x+3+ | | X+3/4x=50 | | 4u-12=48 | | (v+140)/2=420 | | (5z-1)(5+z)=0 | | 5x-4=(2x+1) | | (10i)(8i)=80i | | 62.6-32=1.8c | | 13-3y=5 | | -t^+8t=-48 | | 62=1.6x | | (90-c)/10=4 | | 8t-t^=-48 | | X(4-2y)=3y-1 | | w6=84 | | (v-7)/3=13 | | (x-3)^5=-1024 | | 4x+2(×+3)=2(×+2) |