If it's not what You are looking for type in the equation solver your own equation and let us solve it.
y^2+5y-35=0
a = 1; b = 5; c = -35;
Δ = b2-4ac
Δ = 52-4·1·(-35)
Δ = 165
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{165}}{2*1}=\frac{-5-\sqrt{165}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{165}}{2*1}=\frac{-5+\sqrt{165}}{2} $
| 3×(2m-5)=12 | | 24-3h+14=-7 | | 3-2d=17 | | 3-2d=117 | | 3-x15=24 | | 3+x+15=24 | | x×21=5x+21 | | 3x-12=9+8x | | (1/5)^x-4=1/125 | | (1/5)^x-x=1/125 | | 50+50-25x0+2=2 | | 50+50+25*0+2+2=x | | 3^2x-2.3^x=3 | | 2^x÷3+2^x÷2+1=24 | | 2^y+3+2^y=9 | | 9^(2x-4)=81^(×+3) | | y=400+120 | | 20n-18=42 | | 18m+36=756 | | r=72r | | x-8/9=4 | | (16^1÷2)÷4^3=4^x | | (16^1÷2)÷4^3=4^n | | 34-19k=23 | | 5+7(y-1)=33 | | 8a-2=5a-17 | | 9t^2-15t-6=0 | | m/3+5/2=-3/2 | | 7x-5=15;x=3 | | -(-6-3x)=15 | | 13m=m+1 | | X+5×x-2=60 |